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Powered by Question2Answerugc net generl paper-1
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The value of X. which is exceed 10% of the time in the duration of mesurment isStatistics Answershttps://www.mathhomeworkanswers.org/261861/ugc-net-generl-paper-1Mon, 03 Jun 2019 10:47:19 +0000Answered: A dairy farmer claims that his milk bottles contain exactly 1l of milk. A consumer took a random sample of 20 bottles & found the average contents to be 0.980l with s.d of 0.07l. Test the claim at 5%.
https://www.mathhomeworkanswers.org/246059/bottles-contain-exactly-consumer-bottles-average-contents?show=261842#a261842
<p style="text-align: justify;">The null hypothesis H0: mean=1 litre</p>
<p style="text-align: justify;">The alternative hypothesis Ha: mean≠1 litre</p>
<p style="text-align: justify;">Significance level ɑ=0.05 (2-tail), so we use ɑ/2 for each tail.</p>
<p style="text-align: justify;">Test statistic T=(0.98-1)/(0.07/√20)=-1.278.</p>
<p style="text-align: justify;">The critical value for a 2-tail t-test with 20-1=19 dof is 2.093. We compare -1.278 with left tail -2.093, and since the test statistic is greater than the critical value we conclude that there is insufficient evidence to reject H0, with a 95% degree of confidence. We cannot reject the farmer’s claim.</p>Statistics Answershttps://www.mathhomeworkanswers.org/246059/bottles-contain-exactly-consumer-bottles-average-contents?show=261842#a261842Sat, 01 Jun 2019 18:57:26 +0000Answered: the average of of initial public offering prce for stocks in 1999 was $18 per share. at 1% significant level, conduct a hypothesis test to determine if that the mean offer price was $18 per share
https://www.mathhomeworkanswers.org/252128/initial-offering-significant-conduct-hypothesis-determine?show=261810#a261810
<p style="text-align:justify">The sample statistics give a mean of $17.894 and a standard deviation of $6.011. So (2) x bar=$17.894 and s=$6.011. Sample size n=15. (3) Normal distribution is assumed since median ($16) is fairly close the mean ($17.894). The outlier $34 has skewed the distribution a little so that the mean>median.</p>
<p style="text-align:justify">(1) H0: µ=$18</p>
<p style="text-align:justify">Ha: µ≠$18</p>
<p style="text-align:justify">Significance level ɑ=0.01 (2-tail), so ɑ/2=0.005.</p>
<p style="text-align:justify">(4) The box plot shows that the distribution is positively skewed because of the outlier $34. This outlier has shifted the mean to the right of the median.</p>
<p style="text-align:justify"><img alt="" src="https://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=6764978807213681750" style="height:109px; width:400px"></p>
<p style="text-align:justify">Test statistic T=(17.894-18)/(6.011/√15)=-0.0683. (5) This corresponds to a probability p-value of 0.473, only a little below the mean and much greater than the significance level of 0.005.</p>
<p style="text-align:justify">(6) We have insufficient evidence to refute the claim that the average public offering is $18 per unit.</p>Statistics Answershttps://www.mathhomeworkanswers.org/252128/initial-offering-significant-conduct-hypothesis-determine?show=261810#a261810Thu, 30 May 2019 14:00:00 +0000Answered: A simple random sample of the sitting heights of 36 male students has a mean of 92.8cm. The population of males has...
https://www.mathhomeworkanswers.org/261534/simple-random-sample-sitting-heights-students-population?show=261776#a261776
<p style="text-align:justify">SAMPLE STATISTICS: mean sitting height (x bar)=92.8cm, size (n)=36 male students.</p>
<p style="text-align:justify">POPULATION STATISTIC: mean sitting height (µ)=91.4cm, standard deviation (σ)=3.6cm.</p>
<p style="text-align:justify">CLAIM: Mean sitting height is different from 91.4cm.</p>
<p style="text-align:justify">COUNTERCLAIM: Mean sitting height is 91.4cm.</p>
<p style="text-align:justify">SIGNIFICANCE LEVEL: ɑ=0.05 (95% confidence level)</p>
<p style="text-align:justify">Null hypothesis, H₀: µ=91.4 (counterclaim)</p>
<p style="text-align:justify">Alternative hypothesis, H₁: µ≠91.4 (claim)</p>
<p style="text-align:justify">This is a 2-tail test, because “not equals” means less than (left tail) or greater than (right tail), so the significance level is split equally between the two tails giving us the required confidence interval of 95%.</p>
<p style="text-align:justify">TEST STATISTIC:</p>
<p style="text-align:justify">Z=(x bar-µ)/(σ/√n)=(92.8-91.4)/(3.6/√36)=1.4/0.6=2.333, corresponding to a P-value of 0.0098. The given test statistic is 0.0198.</p>
<p style="text-align:justify">ɑ/2=0.025 and sample P-value 0.0098<0.025, and 0.0198<0.025, which means that H₀ is rejected, and H₁ is accepted to be true. The conclusion is that the claim is true, that the mean sitting height of male students is different from 91.4cm.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261534/simple-random-sample-sitting-heights-students-population?show=261776#a261776Tue, 28 May 2019 16:18:58 +0000Answered: Eighty-one random people were surveyed about the time it takes to commute to work in the morning. The standard deviation of...
https://www.mathhomeworkanswers.org/261535/eighty-random-surveyed-commute-morning-standard-deviation?show=261774#a261774
<p style="text-align:justify">For a 2-tail test (which this seems to be), with 80 dof and ɑ=0.1, the critical value from t-tables is 1.664.</p>
<p style="text-align:justify">If the T value is (2.3-2)/(2.3/√81)=1.17 as the test statistic, then 1.17<1.664, so the null hypothesis would be rejected, which means that the claim is incorrect and the standard deviation is not 2.0 minutes.</p>
<p style="text-align:justify">Initial conclusion may be that from the sample result, the population standard deviation is not 2.0 minutes, because the sample has SD=2.3 minutes.</p>
<p style="text-align:justify">The given test statistic of 105.8 doesn’t seem to tally with the other data. Perhaps it’s the mean commute time?</p>Statistics Answershttps://www.mathhomeworkanswers.org/261535/eighty-random-surveyed-commute-morning-standard-deviation?show=261774#a261774Tue, 28 May 2019 13:08:03 +0000Answered: A watch designer claims that men have wrist breadths with a mean equal to 9 cm. A simple random sample of wrist...
https://www.mathhomeworkanswers.org/261530/watch-designer-claims-breadths-simple-random-sample-wrist?show=261773#a261773
<p style="text-align:justify">SAMPLE STATISTICS: mean wrist breadth (x bar)=8.91cm, size (n)=72 men.</p>
<p style="text-align:justify">POPULATION STATISTIC: standard deviation (σ)=0.36cm. </p>
<p style="text-align:justify">CLAIM: Men have mean wrist breadths=9cm.</p>
<p style="text-align:justify">COUNTERCLAIM: Men have mean wrist breadths not equal to 9cm.</p>
<p style="text-align:justify">SIGNIFICANCE LEVEL: ɑ=0.01 (99% confidence level)</p>
<p style="text-align:justify">Null hypothesis, H₀: µ=9 (claim)</p>
<p style="text-align:justify">Alternative hypothesis, H₁: µ≠9(counterclaim)</p>
<p style="text-align:justify">This is a 2-tail test.</p>
<p style="text-align:justify">TEST STATISTIC: Z=(x bar-µ)/(σ/√n)=(8.91-9)/(0.36/√72)=-2.121, corresponding to a P-value of 0.017. Because we have a 2-tail test we compare this value with 0.005 because the tail probability is divided equally by the two tails. Since 0.017>0.005, we fail to reject the null hypothesis, so we have insufficient evidence to support or counter the claim.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261530/watch-designer-claims-breadths-simple-random-sample-wrist?show=261773#a261773Tue, 28 May 2019 12:10:49 +0000Answered: A simple random sample of 20 women’s hair lengths prior to entering a local hair salon is found to have a standard deviation of 1.2 inches. Use a 0.01 significance level to...
https://www.mathhomeworkanswers.org/261536/random-lengths-entering-standard-deviation-significance?show=261771#a261771
<p style="text-align: justify;">To find the critical value we need to know 3 things, degrees of freedom, significance or confidence level, whether 1- or 2-tail test.</p>
<p style="text-align: justify;">First, what is the claim: σ (population standard deviation) > 1 inch, and the counterclaim is σ≤1.</p>
<p style="text-align: justify;">The null hypothesis, H0: σ=1 (based on the counterclaim which includes σ=1 as well as σ<1); the alternative hypothesis, H1: σ>1 (claim). This is a right tail test (1-tail) because the right tail of the distribution deals with standard deviations above the mean, that is, greater than.</p>
<p style="text-align: justify;">The critical value is found from a t-table because we only know the sample standard deviation, s. We have dof=19, ɑ=0.01, 1-tail test, and the critical value from the table is 2.54. Now we need our test statistic:</p>
<p style="text-align: justify;">(s-σ)/(s/√n)=(1.2-1)/(1.2/√20)=0.745. Since 0.745<2.54, we fail to reject H0, which means we have insufficient evidence to support the claim or counterclaim.</p>
<p style="text-align: justify;"> </p>Statistics Answershttps://www.mathhomeworkanswers.org/261536/random-lengths-entering-standard-deviation-significance?show=261771#a261771Tue, 28 May 2019 11:18:31 +0000Answered: The mean of a sample size n = 35 is 1860. The standard deviation of the sample is 102 and the population is...
https://www.mathhomeworkanswers.org/261520/mean-sample-size-1860-standard-deviation-sample-population?show=261767#a261767
<p style="text-align: justify;">First we calculate the test statistic, which we do using a t-table because we don’t know the population standard deviation. This table needs the number of degrees of freedom, n-1=34.</p>
<p style="text-align: justify;">T=(1860-µ)/(102/√35) where µ is the population mean. Look up T for 34 dof and a 2-tail test at a confidence level of 99% (significance level 100-99=1% or 0.01). T=2.73 approx. So µ is defined by a range, T=±2.73, and µ=1860±2.73(102/√35)=1860±47, so 1813<µ<1907.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261520/mean-sample-size-1860-standard-deviation-sample-population?show=261767#a261767Tue, 28 May 2019 10:36:01 +0000Answered: A sample size of n = 20 is a simple random sample selected from a normally distributed population. Find the critical value ta/2 corresponding to a 95% confidence level.
https://www.mathhomeworkanswers.org/261519/normally-distributed-population-corresponding-confidence?show=261765#a261765
<p><img alt="" src="https://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=354618010693192064" style="height:259px; width:400px"></p>
<p style="text-align:justify">The above picture shows you what you need to look up. The degrees of freedom is one less than the sample size. Significance level is 1-confidence level=1-0.95=0.05. t(ɑ/2) signifies 2-tail because half the significance level is in the left tail of the normal distribution and half in the right tail.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261519/normally-distributed-population-corresponding-confidence?show=261765#a261765Tue, 28 May 2019 10:14:26 +0000Answered: The claim that 90% of Americans check their mail daily is tested. Also assume that the initial conclusion rejects the null hypothesis. State the final conclusion in simple, nontechnical terms
https://www.mathhomeworkanswers.org/261438/americans-conclusion-hypothesis-conclusion-nontechnical?show=261762#a261762
<p style="text-align:justify">The statistical method would be to conduct a simple survey, taking a random sample of Americans and asking the simple yes-no question: “Do you check your mail every day?” So the results would just be so many yesses and so many nos, a simple proportion. The stated claim becomes the null hypothesis for the purpose of testing, that is, 90%, or 0.9, of the results of the survey should be yes. The alternative hypothesis is that the figure differs from 90%.</p>
<p style="text-align:justify">The survey finds that the figure is not 90% and the initial conclusion is to reject the claim. But how confident can we be that our conclusion is representative of all Americans? So a confidence level is set, for example, 95% certainty. From this confidence level we can say that we can expect detailed analysis to show that there is a less than 5% probability that our results are due to chance. After analysis this is what’s found and this gives us confidence that we can reject the claim (the null hypothesis) and establish the counterclaim (the alternative hypothesis) that the figure of 90% is incorrect.</p>
<p style="text-align:justify">To publish the result of the findings of the analysis, we state that we (are 95% certain that we) have sufficient evidence (the sample data) to reject the claim (null hypothesis) that 90% of Americans check their mail every day.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261438/americans-conclusion-hypothesis-conclusion-nontechnical?show=261762#a261762Tue, 28 May 2019 09:10:00 +0000Answered: The standard deviation of sample size of n = 15 is 345 and the mean is 6784. Given that the population follows a...
https://www.mathhomeworkanswers.org/261518/standard-deviation-sample-size-mean-given-population-follows?show=261755#a261755
<p style="text-align:justify"><img alt="" src="https://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=2848938652577906731" style="height:310px; width:400px"></p>
<p style="text-align:justify">The picture shows the normal distribution curve with Z values along the horizontal axis. Z=0 represents the mean which splits the area under the curve into two equal parts. On the left are negative values for Z and on the right are positive values. The area under the curve represents probabilities. The two vertical dotted lines split the areas. The left dotted line is at Z=-1.645 and the right at Z=1.645. The area of the curve between the two lines represents the confidence interval of 90% with 5% in each of the two tails. This picture should help to make sense of this question and its solution.</p>
<p style="text-align:justify">The Z-score is calculated from Z=(x bar-µ)/(σ/√n), where µ and σ are the population mean and standard deviation. x bar=6784, n=15. We can rewrite the equation to find µ in terms of the other values:</p>
<p style="text-align:justify">µ=x bar-Zσ/√n, and we have two values for Z on either side of the mean, so we get two values for µ and this is where we get the range of values for µ. But we don’t have the population standard deviation, we have the sample standard deviation, s, so we need to consult a t-table to find a critical value for n-1=14 degrees of freedom and a 2-tail significance of 0.10. That value is T=1.761 and we use this in place of Z: Ts/√15=157 approx. Therefore we have the range for µ: 6784±157=(6627,6941), in other words 6627<µ<6941.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261518/standard-deviation-sample-size-mean-given-population-follows?show=261755#a261755Tue, 28 May 2019 00:38:26 +0000Answered: A simple random sample has a sample size of n = 65. Given the population is normally distributed, find the critical value...
https://www.mathhomeworkanswers.org/261521/random-sample-sample-population-normally-distributed-critical?show=261754#a261754
<p style="text-align:justify">In your t-table look for the row containg 64 degrees of freedom because dof=n-1. Find the 2-tail column for 0.01, at the row-column intersection you find the figure 2.660 (or approximately so depending on the accuracy and scope of your table).</p>
<p style="text-align:justify">We use 2-tail figures t(ɑ/2) when we’re testing for equal/not equal, otherwise we want 1-tail—the left tail of the normal distribution for less than (or less than or equal) or the right tail for greater than (or greater than or equal). That’s why the columns have two different labels, depending on what condition you are testing.</p>
<p style="text-align:justify">The significance level ɑ and the confidence level are related. 0.01 significance (1%) is the same as 0.99 (99%) confidence level. The two always add up to 1 (100%).</p>Statistics Answershttps://www.mathhomeworkanswers.org/261521/random-sample-sample-population-normally-distributed-critical?show=261754#a261754Mon, 27 May 2019 23:46:28 +0000Answered: The totals of the individual weights of garbage discarded by 62 households in one week have a mean of 27.553 lb. Assume...
https://www.mathhomeworkanswers.org/261528/totals-individual-weights-garbage-discarded-households-assume?show=261753#a261753
<p style="text-align:justify">SAMPLE STATISTICS: mean weight of garbage (x bar)=27.553lb, size (n)=62 households.</p>
<p style="text-align:justify">POPULATION STATISTIC: standard deviation (σ)=12.458lb. </p>
<p style="text-align:justify">CLAIM: Mean weight of garbage is less than 30lb.</p>
<p style="text-align:justify">COUNTERCLAIM: Mean weight of garbage is greater than or equal to 30lb.</p>
<p style="text-align:justify">SIGNIFICANCE LEVEL: ɑ=0.05 (95% confidence level)</p>
<p style="text-align:justify">Null hypothesis, H₀: µ=30 (from counterclaim, which permits µ=30)</p>
<p style="text-align:justify">Alternative hypothesis, H₁: µ<30 (claim)</p>
<p style="text-align:justify">This is a left tail test.</p>
<p style="text-align:justify">TEST STATISTIC: Z=(x bar-µ)/(σ/√n)=(27.553-30)/(12.458/√62)=-1.547, corresponding to a P-value of 0.061.</p>
<p style="text-align:justify">ɑ=0.05 and sample P-value 0.061>0.05, which means that H₀ fails to be rejected. We conclude that there is insufficient evidence to decide with 95% confidence whether or not the mean weight of garbage is less than 30lb. Although the sample weight was less than 30lb, it wasn’t significantly less.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261528/totals-individual-weights-garbage-discarded-households-assume?show=261753#a261753Mon, 27 May 2019 23:09:24 +0000Answered: Find the chi-square value X2/L corresponding to a sample size of 17 and a confidence level of 98%
https://www.mathhomeworkanswers.org/261514/find-square-value-corresponding-sample-size-confidence-level?show=261748#a261748
<p style="text-align: justify;">To find lower chi-squared value when the table gives values to the right of the significance level we need to use 100-98=2% significance level, ɑ=0.02 for dof 17-1=16: 29.63.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261514/find-square-value-corresponding-sample-size-confidence-level?show=261748#a261748Mon, 27 May 2019 20:09:34 +0000Answered: A simple random sample of 36 bottles of Helo, a healthy juice mix, has a mean volume of 12.19 oz. Assume that the...
https://www.mathhomeworkanswers.org/261529/simple-random-sample-bottles-healthy-juice-volume-assume-that?show=261746#a261746
<p style="text-align:justify">SAMPLE STATISTICS: mean volume (x bar)=12.19 fluid oz, size (n)=36 bottles.</p>
<p style="text-align:justify">POPULATION STATISTIC: standard deviation (σ)=0.11 fluid oz</p>
<p style="text-align:justify">CLAIM: Mean volume is 12 fluid oz</p>
<p style="text-align:justify">COUNTERCLAIM: Mean volume is not 12 fluid oz</p>
<p style="text-align:justify">SIGNIFICANCE LEVEL: ɑ=0.01 (99% confidence level)</p>
<p style="text-align:justify">Null hypothesis, H₀: µ=12 (claim)</p>
<p style="text-align:justify">Alternative hypothesis, H₁: µ≠12 (counterclaim)</p>
<p style="text-align:justify">This is a 2-tail test, because “not equals” means less than (left tail) or greater than (right tail), so the significance level is split between the two tails giving us the required confidence interval of 99%.</p>
<p style="text-align:justify">TEST STATISTIC: Z=(x bar-µ)/(σ/√n)=(12.19-12)/(0.11/√36)=10.37, corresponding to a P-value of 0.000. ɑ/2=0.005 and sample P-value 0.000<0.005, which means that H₀ is rejected, and H₁ is accepted to be true. The conclusion is that the stated claim is false, and the mean volume of Helo is not 12 oz.</p>
<p style="text-align:justify">Another way of confirming this result is to use the 2-tail critical value for the 0.01 significance level, which corresponds to |Z|=2.575. 10.37>2.575 meaning that the sample Z-score is more extreme (way more extreme!) than the critical value, and H₀ is rejected and H₁ is accepted.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261529/simple-random-sample-bottles-healthy-juice-volume-assume-that?show=261746#a261746Mon, 27 May 2019 19:41:26 +0000Answered: A recent claim stated the mean temperature setting at a person’s home is 78.3 degrees F. A simple random sample of...
https://www.mathhomeworkanswers.org/261532/stated-temperature-setting-persons-degrees-random-sample?show=261743#a261743
<p style="text-align: justify;">SAMPLE STATISTICS: mean home temperature setting (x bar)=77.5°F, size (n)=40 homes.</p>
<p style="text-align: justify;">POPULATION STATISTIC: standard deviation (σ)=1.9°F</p>
<p style="text-align: justify;">CLAIM: Mean home temperature setting is 78.3°F.</p>
<p style="text-align: justify;">COUNTERCLAIM: Mean home temperature setting is not 78.3°F.</p>
<p style="text-align: justify;">SIGNIFICANCE LEVEL: ɑ=0.05 (95% confidence level)</p>
<p style="text-align: justify;">Null hypothesis, H₀: µ=78.3°F(claim)</p>
<p style="text-align: justify;">Alternative hypothesis, H₁: µ≠78.3°F (counterclaim)</p>
<p style="text-align: justify;">This is a 2-tail test, because “not equals” means less than (left tail) or greater than (right tail), so the significance level is split between the two tails giving us the required confidence interval of 95%.</p>
<p style="text-align: justify;">TEST STATISTIC: Z=(x bar-µ)/(σ/√n)=(77.5-78.3)/(1.9/√40)=-2.66, corresponding to a P-value of 0.0039.</p>
<p style="text-align: justify;">ɑ/2=0.025 and sample P-value 0.0039<0.025, which means that H₀ is rejected, and H₁ is accepted to be true. The conclusion is the claim that the mean home temperature is not 78.3°F.</p>
<p style="text-align: justify;">Another way of confirming this result is to use the 2-tail critical value for the 0.05 significance level, which corresponds to |Z|=1.96. -2.66 < -1.96, meaning that the sample Z-score is more extreme than the critical value, and H₀ is rejected and H₁ is accepted.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261532/stated-temperature-setting-persons-degrees-random-sample?show=261743#a261743Mon, 27 May 2019 18:47:21 +0000Answered: A recent survey was conducted with a simple random sample of 35 shoppers at the food court tin a local mall. The mall claims that...
https://www.mathhomeworkanswers.org/261533/recent-survey-conducted-simple-random-sample-shoppers-claims?show=261742#a261742
<p style="text-align: justify;">SAMPLE STATISTICS: mean price of lunch (x bar)=$4.49, size (n)=35 shoppers.</p>
<p style="text-align: justify;">POPULATION STATISTIC: standard deviation (σ)=$0.36.</p>
<p style="text-align: justify;">CLAIM: Mean price of lunch is less than $4.65.</p>
<p style="text-align: justify;">COUNTERCLAIM: Mean price of lunch is greater than or equal to $4.65.</p>
<p style="text-align: justify;">Null hypothesis, H₀: µ=$4.65 (from the counterclaim, because greater than or equal specifically includes the equality)</p>
<p style="text-align: justify;">Alternative hypothesis, H₁: µ<$4.65 (the claim)</p>
<p style="text-align: justify;">SIGNIFICANCE LEVEL: ɑ=0.01 (99% confidence level), left tail</p>
<p style="text-align: justify;">TEST STATISTIC: Z=(x bar-µ)/(σ/√n)=(4.49-4.65)/(0.36/√35)=-2.63, which does indeed correspond to a P-value of 0.0043.</p>
<p style="text-align: justify;">ɑ=0.01 (99% confidence level) and sample P-value 0.0043<0.01, which means that H₀ is rejected, and H₁ is accepted to be true. The conclusion is the claim that the mean price of lunch is less than $4.65.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261533/recent-survey-conducted-simple-random-sample-shoppers-claims?show=261742#a261742Mon, 27 May 2019 18:19:13 +0000Answered: A diet plan claims that women have a mean weight of 145 lbs. Assume that the weights of women have a standard deviation...
https://www.mathhomeworkanswers.org/261531/claims-women-weight-assume-weights-women-standard-deviation?show=261741#a261741
<p style="text-align:justify">SAMPLE STATISTICS: mean (x bar)=153.2lb, size (n)=40 women.</p>
<p style="text-align:justify">POPULATION STATISTIC: standard deviation (σ)=30.86lb, mean (µ)=145lb.</p>
<p style="text-align:justify">CLAIM: Women have a mean weight of 145lbs.</p>
<p style="text-align:justify">COUNTERCLAIM: The mean weight of women is not equal to 145lbs.</p>
<p style="text-align:justify">Null hypothesis, H₀: µ=145lbs</p>
<p style="text-align:justify">Alternative hypothesis, H₁: µ≠145lbs</p>
<p style="text-align:justify">SIGNIFICANCE LEVEL: ɑ=0.05 (95% confidence level), 2-tail.</p>
<p style="text-align:justify">TEST STATISTIC: Z=(x bar-µ)/(σ/√n)=(153.2-145)/(30.86/√40),</p>
<p style="text-align:justify">Z=8.2/4.88=1.68.</p>
<p style="text-align:justify">To find p-value, look up Z=-1.68 or look up Z=1.68 and subtract from 1.</p>
<p style="text-align:justify">p-value=0.0465.</p>
<p style="text-align:justify">We have a 2-tail test because H₁ implies that the pop mean could be either less than or greater than 145lb. The significance level has to be split between the left and right tails of the distribution, so we compare the p-value of the test statistic with 0.05/2=0.025. Since 0.0465>0.025 we are within the 95% confidence level so we have insufficient evidence within the sample data to reject the null hypothesis. That is, the sample data do not differ significantly from the expected mean of 145lb.</p>
<p style="text-align:justify">(Note that if H₁ had been >145lb, a right-tail test would have applied and since 0.0465≤0.05 the null hypothesis would have been rejected, and by default H₁ would have applied.)</p>Statistics Answershttps://www.mathhomeworkanswers.org/261531/claims-women-weight-assume-weights-women-standard-deviation?show=261741#a261741Mon, 27 May 2019 16:35:02 +0000Answered: i NEED HELP WITH NULL HYPOTHESIS
https://www.mathhomeworkanswers.org/185232/i-need-help-with-null-hypothesis?show=261690#a261690
<p style="text-align:justify">H0: p=0.11 (the claim)</p>
<p style="text-align:justify">Ha: p≠0.11 (counterclaim)</p>
<p style="text-align:justify">q=1-p=0.89</p>
<p style="text-align:justify">Sample p hat=37/167=0.2216 approx (22.16%)</p>
<p style="text-align:justify">Sample n=167</p>
<p style="text-align:justify">Test statistic for test: (p hat-p)/√(pq/n)=</p>
<p style="text-align:justify">(0.2216-0.11)/0.0242=0.1116/0.0242=4.607.</p>
<p style="text-align:justify">The probability associated with this value is about 0.000 (p-value) from which we conclude (the critical probability is 0.025) that we reject H0 and default to Ha, the proportion is not 11%.</p>
<p style="text-align:justify"> </p>Statistics Answershttps://www.mathhomeworkanswers.org/185232/i-need-help-with-null-hypothesis?show=261690#a261690Fri, 24 May 2019 19:32:17 +0000Answered: hypothesis testing
https://www.mathhomeworkanswers.org/74227/hypothesis-testing?show=261689#a261689
<p style="text-align:justify">H0: p=0.06 (the growers’ claim)</p>
<p style="text-align:justify">Ha: p≠0.06</p>
<p style="text-align:justify">Type of test: proportion, 2-tailed</p>
<p style="text-align:justify">Significance: 5%=0.05, but this is split between the two tails: 0.025 for each tail corresponds to a Z-score of 1.96.</p>
<p style="text-align:justify">Test statistic for proportions: (0.8-0.6)/√(pq/1000), where q=1-p=0.94.</p>
<p style="text-align:justify">Z-score for the test statistic=0.2/0.00751=26.63. This is an extremely large value corresponding to a probability of almost zero. H0 is therefore rejected. We therefore state that the proportion is not 6%.</p>Statistics Answershttps://www.mathhomeworkanswers.org/74227/hypothesis-testing?show=261689#a261689Fri, 24 May 2019 19:10:17 +0000Answered: A simple random sample of 28 wait times while pumping gas at gas station has a standard deviation of 20 seconds. The test statistic for...
https://www.mathhomeworkanswers.org/261538/random-pumping-station-standard-deviation-seconds-statistic?show=261685#a261685
<p style="text-align: justify;">The t value for the test statistic is (22.314-22)/(20/√28)=0.0831 approx. The number of degrees of freedom is 28-1=27. The critical value for a significance level of 0.05 for this is 1.703 for a 1-tail test (because we are testing an inequality standard deviation<22 seconds, the null hypothesis claim). Null hypothesis is the stated claim; the alternative hypothesis is that the standard deviation is greater than or equal to 22 seconds. 0.0831 is less than the critical value, so we reject the null hypothesis and by default conclude that the standard deviation is greater than or equal to 22 seconds with a confidence of 95%.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261538/random-pumping-station-standard-deviation-seconds-statistic?show=261685#a261685Fri, 24 May 2019 12:34:18 +0000Answered: 1. Given S = {x│x is a distinct letter of the word “prism”}. How many valid events of the sample space S can be formed if it contains “s” but does not contain “m”?
https://www.mathhomeworkanswers.org/261618/distinct-letter-valid-events-sample-formed-contains-contain?show=261646#a261646
We can eliminate “m”, and we are left with “s” and its permutations with “p”, “r” and “i”. This gives us:<br />
<br />
s; sp, sr, si, ps, rs, is; spr, spi, sir, and all the permutations of these; spri and all its permutations.<br />
<br />
There are 6 each of the 3-letter permutations.<br />
<br />
When we add all these together we get: 1+6+3×6+24=1+6+18+24=49.Statistics Answershttps://www.mathhomeworkanswers.org/261618/distinct-letter-valid-events-sample-formed-contains-contain?show=261646#a261646Thu, 23 May 2019 16:23:45 +0000Answered: A shipment of 12 television sets contains 3 defective sets. In how many ways can a hotel purchase 5 of these sets and receive at least 2 of the defective sets?
https://www.mathhomeworkanswers.org/261620/shipment-television-contains-defective-purchase-defective?show=261643#a261643
<p style="text-align:justify">Two separate sets: 9 “good” TVs and 3 “bad” TVs make up 12 TV sets.</p>
<p style="text-align:justify">The 5 TVs delivered to the hotel must consist of at least 2 bad TVs. So that could be 3 bad TVs and 2 good ones; or 3 good TVs and 2 bad ones.</p>
<p style="text-align:justify">There is only one way to get 3 bad TVs, because there are only 3 of them anyway. But there are 9×8/2=36 ways of getting 2 good TVs out of 9.</p>
<p style="text-align:justify">There are 3 ways of getting 2 bad TVs out of 3, and 9×8×7/(2×3)=84 ways of getting 3 good TVs out of 9. Combine these combinations: 3×84=252.</p>
<p style="text-align:justify">Add these combinations together (2 bad/3 good+2 good/3 bad): 252+36=288 ways of getting at least 2 defective TVs.</p>
<p style="text-align:justify">(If this were to be converted to a probability, we’d divide 288 by the number of ways 5 TVs could be selected out of 12, 288/792=4/11.)</p>Statistics Answershttps://www.mathhomeworkanswers.org/261620/shipment-television-contains-defective-purchase-defective?show=261643#a261643Thu, 23 May 2019 12:58:10 +0000Answered: In how many ways can 6 people be arranged in a round table if a certain 3 persons refuse to follow each other?
https://www.mathhomeworkanswers.org/261619/people-arranged-table-certain-persons-refuse-follow-other?show=261641#a261641
<p style="text-align:justify">Represent the table as a clock with people seated at 2, 4, 6, 8, 10, 12 hours.</p>
<p style="text-align:justify">Let the people be represented as letters A-F. Let also A, B and C identify those people who do not wish to sit together and put them in table positions 2, 6 and 10, so that D, E and F sit in positions 4, 8 and 12. A, B and C can shift positions in 6 different ways, and so can D, E and F. For every position of A, B and C, there are 6 different positions for D, E and F, therefore there are 36 different arrangements. But each unique arrangement of people can start at a different seat: 2, 4, 6, 8, 10, 12, making the total number of arrangements=6×36=216.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261619/people-arranged-table-certain-persons-refuse-follow-other?show=261641#a261641Thu, 23 May 2019 11:43:44 +0000Answered: Three dice are thrown. What is the probability that the same number appears on exactly two of the dice?
https://www.mathhomeworkanswers.org/261622/three-dice-thrown-probability-same-number-appears-exactly?show=261639#a261639
<p style="text-align:justify">Let the letters a, b, c, d, e, f represent the numbers 1-6, but not necessarily a=1, b=2, etc. The three dice are thrown. The first shows value a. There are 10 ways to get another a (and no more):</p>
<p style="text-align:justify">aab, aac, aad, aae, aaf, aba, aca, ada, aea, afa.</p>
<p style="text-align:justify">And there are another 5 ways where the other two dice duplicate their values:</p>
<p style="text-align:justify">abb, acc, add, aee, aff.</p>
<p style="text-align:justify">So that’s 15 duplications out of a possible 36 combinations of the second and third dice, where a is the value of the first die.</p>
<p style="text-align:justify">What applies to a symmetrically applies to the other values b-f. So the probability is 15/36=5/12. (Or we could simply state this as 90/216 for all possible outcomes. That is, out of the 216 possible outcomes for throwing three dice, 90 of them contain exactly one pair of duplicated values.)</p>Statistics Answershttps://www.mathhomeworkanswers.org/261622/three-dice-thrown-probability-same-number-appears-exactly?show=261639#a261639Thu, 23 May 2019 11:10:33 +0000Answered: While dressing in the dark you select 2 socks from a drawer containing 5 differently colored pairs. What is the probability that your socks match?
https://www.mathhomeworkanswers.org/261623/dressing-drawer-containing-differently-colored-probability?show=261635#a261635
<p style="text-align: justify;">There are 10 socks in the drawer. You take one sock out, leaving 9 others, only one of which matches the sock you took out, so there is 1/9 chance you will pick it out next.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261623/dressing-drawer-containing-differently-colored-probability?show=261635#a261635Thu, 23 May 2019 10:04:54 +0000Answered: A fair die is rolled until a 4 appears. What is the probability that this die must be rolled more than 10 times?
https://www.mathhomeworkanswers.org/261621/rolled-until-appears-probability-this-rolled-more-than-times?show=261634#a261634
<p style="text-align: justify;">The probability of getting a 4 on one roll of the die is 1/6.</p>
<p style="text-align: justify;">The probability of having to roll the die twice to get a 4 is 5/6×1/6=5/36, because there is a 5/6 probability of failing to get a 4 on the first roll and a 1/6 probability of getting 4 on the second roll.</p>
<p style="text-align: justify;">Failing to get 4 on 10 rolls is (5/6)¹⁰ but getting 4 on the next roll is (5/6)¹⁰(1/6)=5¹⁰/6¹¹=0.02692 approx. Let P=5¹⁰/6¹¹.</p>
<p style="text-align: justify;">Not getting 4 until the 12th roll is (5/6)¹¹(1/6)=0.02243=P×5/6, and so on. We need to sum these probabilities from 11 to infinity:</p>
<p style="text-align: justify;">P(1+5/6+(5/6)²+(5/6)³+...)=P/(1-(5/6))=6P=(5/6)¹⁰=0.1615 approx, 16.15%.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261621/rolled-until-appears-probability-this-rolled-more-than-times?show=261634#a261634Thu, 23 May 2019 09:58:19 +0000Answered: 11. A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 4 times, what is the probability of getting a. exactly 2 tails? b. at least 3 heads?
https://www.mathhomeworkanswers.org/261624/biased-likely-occur-tossed-probability-getting-exactly-heads?show=261630#a261630
<p style="text-align: justify;"> Probability of H=2/3, probability of T=1/3.</p>
<p style="text-align: justify;">All 16 outcomes with their probabilities:</p>
<p style="text-align: justify;">HHHH=16/81</p>
<p style="text-align: justify;">HHHT, HHTH, HTHH, THHH=8/81</p>
<p style="text-align: justify;">HHTT,HTHT,HTTH,THHT,THTH,TTHH=4/81</p>
<p style="text-align: justify;">HTTT,THTT,TTHT,TTTH=2/81</p>
<p style="text-align: justify;">TTTT=1/81</p>
<p style="text-align: justify;">a) Probability of exactly 2 tails: 6×4/81=24/81=8/27.</p>
<p style="text-align: justify;">b) Probability of at least 3 heads: 4×8/81+16/81=48/81=16/27.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261624/biased-likely-occur-tossed-probability-getting-exactly-heads?show=261630#a261630Thu, 23 May 2019 09:30:02 +0000Answered: solve: units of oxygen produced = age2 - 5 x age - 50
https://www.mathhomeworkanswers.org/261609/solve-units-of-oxygen-produced-age2-5-x-age-50?show=261612#a261612
<p style="text-align: justify;">Let variable a represent age: a²-5a-50 is the expression for the model’s oxygen production.</p>
<p style="text-align: justify;">When a²-5a-50=0=(a-10)(a+5), so a=10 years when production of oxygen=0 (because a>0).</p>
<p style="text-align: justify;">The trees live to be 10 years old.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261609/solve-units-of-oxygen-produced-age2-5-x-age-50?show=261612#a261612Wed, 22 May 2019 21:47:35 +0000Answered: A simple random sample of 50 stainless steel metal screws is obtained. The screws have a mean length of 0.73 inches. Assume the population...
https://www.mathhomeworkanswers.org/261527/random-sample-stainless-screws-obtained-assume-population?show=261608#a261608
<p style="text-align:justify">We need to find the Z score for the difference between the sample mean and the expected population mean.</p>
<p style="text-align:justify">The difference is 0.02 inches, which is 0.02/0.012=20/12=5/3=1.667 standard deviations, and this is the Z score. From tables the Z score corresponds to a probability of 0.4522 of the left side of the normal distribution so it leaves a tail of 0.5000-0.4522=0.0478. This is the p-value associated with the screw length of 0.73 inches. If the significance level is 0.05, the confidence level is 95%, but that extends across the whole distribution. Half the distribution is to the left of the mean, which is 95/2=47.5% or 0.475, leaving 0.500-0.475=0.025 in the left tail. The p-value is greater than this, which means we fail to reject the claim that the screws have a mean length of 0.75 inches. And so the p-value supports the claim within the significance level specified.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261527/random-sample-stainless-screws-obtained-assume-population?show=261608#a261608Wed, 22 May 2019 19:15:44 +0000Answered: A simple random sample of 8 women basketball player’s weights have a standard deviation of 7.5 lb. Find the test statistic...
https://www.mathhomeworkanswers.org/261539/basketball-players-weights-standard-deviation-statistic?show=261598#a261598
<p style="text-align: justify;">Standard error=7.5/√8=2.6517lb approx. The critical value for 95% confidence level (0.05 significance level) is 1.895 for 8-1=7 degrees of freedom (1-tail). The margin of error is 1.895×2.6517=5.02lb approx. This is a 1-tailed critical value because we are testing for less than 9lb, not less than or greater than. The upper limit of the margin of error in the sample is 12.52lb, which exceeds 9lb. The lower limit is 2.48lb. So I think it’s uncertain that the standard deviation is less than 9lb.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261539/basketball-players-weights-standard-deviation-statistic?show=261598#a261598Wed, 22 May 2019 11:53:57 +0000Answered: A simple random sample of 16 birth weights to mothers given a special vitamin supplement during pregnancy has standard deviation of 0.959 kg. Test the claim that this...
https://www.mathhomeworkanswers.org/261537/weights-mothers-supplement-pregnancy-standard-deviation?show=261594#a261594
<p style="text-align:justify">Standard error=0.959/√16=0.23975kg.</p>
<p style="text-align:justify">Critical value fot 0.05 significance level with 15 degrees of freedom is 2.131. Apply this to the standard error to get the margin of error: 2.131×0.23975=0.5109 approx.</p>
<p style="text-align:justify">Therefore the range for the standard deviation is 0.959±0.511: 0.448 to 1.470kg. The value 0.470kg is within this range so the claim is valid (95% certain).</p>
<p style="text-align:justify">Initial conclusion might have been that the sample standard deviation differed from the population standard deviation; but the final conclusion is that the population standard deviation is within the margin of error as applied to the sample, so the sample analysis shows that the standard deviation is not significantly different from that of the population.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261537/weights-mothers-supplement-pregnancy-standard-deviation?show=261594#a261594Wed, 22 May 2019 10:44:41 +0000Answered: A simple random sample of 25 vitamin tablets is obtained, and the potassium content of each tablet is measured. The sample has a standard deviation of 3.7 mg. Use a 0.05 significance...
https://www.mathhomeworkanswers.org/261540/obtained-potassium-measured-standard-deviation-significance?show=261551#a261551
<p style="text-align: justify;">The standard error is 3.7/√25=3.7/5=0.74mg. We need the critical value for this sample size and significance level of 0.05 (95% confidence level). We use the 2-tailed value because the test is for equality, which means we test for less than or greater than the claim of 3.2mg. Looking up the critical value in the test table gives us 2.064 (24 degrees of freedom=sample size 25-1).</p>
<p style="text-align: justify;">2.064 times 0.74=1.53 approx for the margin of error: 3.7±1.53 gives us a range of 2.17 to 5.23mg, and 3.2mg lies within this range, so the claim is valid.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261540/obtained-potassium-measured-standard-deviation-significance?show=261551#a261551Tue, 21 May 2019 11:36:39 +0000Answered: You are asked to conduct a simulation and run a hypothesis test at a significance level of 0.01. Based on your simulation, you find that the p-value is 0.012.
https://www.mathhomeworkanswers.org/261454/conduct-simulation-hypothesis-significance-based-simulation?show=261470#a261470
<p style="text-align:justify">0.01 corresponds to a confidence level of 90%, so we would expect our probability to be within this 90% interval. Outside the interval we reject the null hypothesis and, by default, accept the alternative hypothesis. 90% leaves 10% shared by the two tails of the symmetrical distribution. So that’s 5%, or 0.05, for each tail. 0.012 is less than 0.05 so we reject the null hypothesis. This 2-tail test applies to the null hypothesis, when the alternative hypothesis is specifically a not-equal type. A 1-tail test applies when the alternative hypothesis is either greater-than (right-tail) or less-than (left-tail). The p-value of 0.012 is much less than the significance level of 0.10, so, again, the null hypothesis is rejected, implying that the alternative hypothesis is accepted with the given level of significance.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261454/conduct-simulation-hypothesis-significance-based-simulation?show=261470#a261470Sun, 19 May 2019 20:00:28 +0000Answered: A principal simulated a student raffle using 500 trials and recorded the grade level of the winner each time. The results are shown in the table
https://www.mathhomeworkanswers.org/261455/principal-simulated-student-raffle-recorded-winner-results?show=261464#a261464
<p style="text-align: justify;">The sum of these counts is 460, not 500 as would be expected, so we conclude that 40 students are outside the listed grades. Assuming that to be the case, the number of grade 11 winners is 80/500 of the total number of students, that is, 0.16 or 16%. This is the probability that the winner would be a grade 11 student.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261455/principal-simulated-student-raffle-recorded-winner-results?show=261464#a261464Sun, 19 May 2019 11:55:45 +0000Answered: In a recent taste test, 1017 out of 1250 people could not tell the difference between a popular soda’s original flavor and the new calorie-free flavor. The company claims that more than 80%
https://www.mathhomeworkanswers.org/261453/recent-difference-between-popular-original-calorie-company?show=261463#a261463
<p style="text-align: justify;">This seems to be a binomial distribution. There are only two possible outcomes: people can tell the difference; or people can’t tell the difference. The company claim that over 80% can’t tell the difference, so that’s 80% of 1250=1000. If we put p=0.80 then np is the mean=1000 and the variance is np(1-p)=1000(0.20)=200, the square root being the standard deviation=√200=14.14 approx. The Z score for 1017 is (1017-1000)/14.14=17/14.14=1.20.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261453/recent-difference-between-popular-original-calorie-company?show=261463#a261463Sun, 19 May 2019 11:47:12 +0000Answered: What is the probability that it will take a resident of this city between 20 and 40 minutes to travel to work? with a mean of 38.3
https://www.mathhomeworkanswers.org/261444/probability-resident-city-between-minutes-travel-work-with?show=261449#a261449
<p style="text-align:justify">20 mins is 18.3 below the mean, and 40 mins is 1.7 mins above the mean. We don’t know the standard deviation. Let’s assume the SD is roughly the square root of the mean, that is, about 6, then we can work out a probability. -18.3/6=-3.05 and 1.7/6=0.28 approx. We can use these as Z scores, Z1=-3.05 and Z2=0.28.</p>
<p style="text-align:justify">From these we get two probabilities: for Z1, p1=1-N(3.05)=1-0.9989=0.0011. For Z2, p2=0.6103.</p>
<p style="text-align:justify">p2-p1=0.6092, or about 0.61.</p>
<p style="text-align:justify">However, we will get different results if we change the SD. Let SD=8 mins.</p>
<p style="text-align:justify">This time Z1=-2.29 and Z2=0.21. p1=0.011 and p2=0.583, p2-p1=0.57 approx.</p>
<p style="text-align:justify">So we can conclude from these examples that the probability is about 60%.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261444/probability-resident-city-between-minutes-travel-work-with?show=261449#a261449Sun, 19 May 2019 00:40:21 +0000Answered: while testing gas pumps for accuracy across the state, the research team found that out of 6985 pumps, 1299 were inaccurate. use a normal distribution and a known test statistic of z = –2.93.
https://www.mathhomeworkanswers.org/261441/testing-accuracy-research-inaccurate-distribution-statistic?show=261443#a261443
<p style="text-align:justify">Using a normal distribution table look up Z=2.93 and you’ll find it corresponds to a probability of 0.9983, which means that 99.83% of the distribution is within 2.93 standard deviations above the mean. It also means that 100-99.83=0.17% are outside this range. Because of symmetry 0.17% lie in the opposite tail of the distribution, that is, more than 2.93 standard deviations below the mean, represented by Z=-2.93. </p>Statistics Answershttps://www.mathhomeworkanswers.org/261441/testing-accuracy-research-inaccurate-distribution-statistic?show=261443#a261443Sat, 18 May 2019 20:02:36 +0000Answered: assume the normal distribution applies to a scenario where you are given h1 p<0.08 in a test using a significance level of a=0.01. Find the critical z value(s)
https://www.mathhomeworkanswers.org/261439/assume-distribution-applies-scenario-significance-critical?show=261442#a261442
<p style="text-align: justify;">In this problem we’re looking at an inequality. Less than means we are looking at the left side of the normal distribution, rather than looking at both sides of the distribution and considering an interval on either side of the mean. All the Z values to the left of the mean (that is, less than the mean) have negative values. The significance level tells us how far over to the left. We are considering the left tail of the distribution, so 99% of the distribution lies to the right of this tail. Tables usually give the percentage to the left of the Z value. Therefore, if the confidence level is 99% (significance level is 1%, ɑ=0.01), the Z value is about 2.33, meaning 99% of the distribution is 2.33 standard deviations above the mean. By symmetry the left tail and the right tail each represent 1% of the distribution. We need the left tail, so the Z value we need is 2.33 standard deviations below the mean, represented by Z=-2.33.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261439/assume-distribution-applies-scenario-significance-critical?show=261442#a261442Sat, 18 May 2019 19:29:20 +0000Answered: Assume you want to construct a 90% confidence interval from sample of a normally distributed population. The sample size is 37. Find the critical value ta/2.
https://www.mathhomeworkanswers.org/261436/construct-confidence-normally-distributed-population-critical?show=261440#a261440
<p style="text-align:justify"><img alt="" src="https://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=807495642095342997" style="height:176px; width:450px"></p>
<p style="text-align:justify">The above is an extract from a t test table. The last column contains figures for a 2-tailed 90% confidence interval. For this we use ɑ/2=0.05 or 5% being half the difference between 90% and 100%. The distribution curve is symmetrical about the mean so to get the confidence “width” to be 90% of the area beneath the curve, we need 95% of what is above the mean—the right half of the distribution. The first column of the table contains the number of degrees of freedom (dof), which in this case is 37-1=36, one less than the sample size. From the table we see that for dof=30, the critical value is 1.697 and for dof=40 it’s 1.684. If we interpolate linearly between these values we estimate that for dof=36, the value is about 1.689.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261436/construct-confidence-normally-distributed-population-critical?show=261440#a261440Sat, 18 May 2019 18:33:21 +0000Answered: A population has a known standard deviation of 1.27, and the sample space contains 85 values. Find the margin of error needed to create a 99% confidence interval estimate of the mean of the population
https://www.mathhomeworkanswers.org/261407/population-standard-deviation-contains-confidence-population?show=261423#a261423
<p style="text-align:justify">Standard error=1.27/√85=0.13775.</p>
<p style="text-align:justify">The critical value for 99% confidence level (2-tailed) is about 2.575 so the margin of error is 2.575×0.13775= 0.3547 aprox.</p>
<p style="text-align:justify">I used ɑ=1-0.99=0.01 then ɑ/2=0.005 to find the Z value corresponding to 1-0.005=0.9950, which is Z=2.575. That is 2.575 standard deviations from the mean.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261407/population-standard-deviation-contains-confidence-population?show=261423#a261423Sat, 18 May 2019 10:08:37 +0000Answered: the sample space of population being researched contains 58 values and a mean of 318.6.
https://www.mathhomeworkanswers.org/261409/sample-space-population-being-researched-contains-values?show=261421#a261421
<p style="text-align:justify">With a sample size of this magnitude we can use 1.96 as the 95% confidence interval critical value from tables as the multiplier for the standard error, which is 29.2/√58=3.834 approx. The margin of error is therefore 1.96×3.834=7.5149. Therefore we have 318.6±7.515 as the error on the mean: 311.085<µ<326.115.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261409/sample-space-population-being-researched-contains-values?show=261421#a261421Sat, 18 May 2019 09:46:54 +0000Answered: A simple random sample of the weights of 40 women has mean of 146.22 lb.
https://www.mathhomeworkanswers.org/261408/simple-random-sample-of-the-weights-of-40-women-has-mean-of-146?show=261418#a261418
<p style="text-align:justify">In this case we need to work out the standard error first=30.86/√40=4.88 approx. Next we find the 2-tailed critical value for 95% (where significance level ɑ=1-0.95=0.05) and from normal distribution tables this corresponds to 1.96 standard deviations. We multiply 4.88 by this critical value=9.564 approx. This gives us the margin on either side of the mean: 146.22±9.56: 136.66<µ<155.78.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261408/simple-random-sample-of-the-weights-of-40-women-has-mean-of-146?show=261418#a261418Sat, 18 May 2019 08:58:00 +0000Answered: Find the critical value za/2 corresponding to a 99.5% confidence level.
https://www.mathhomeworkanswers.org/261405/find-the-critical-value-za-corresponding-confidence-level?show=261406#a261406
<p style="text-align:justify">ɑ=1-0.995=0.005, so ɑ/2=0.0025. This is the significance level. It corresponds to 1-0.0025=0.9975 confidence level (for a 2-tailed distribution) and from a normal distribution tables Z=2.81 gives us a probability of 0.9975. This means that 99.75% of the distribution (represented by the area under the distribution curve) is 2.81 standard deviations above or below the mean, giving a confidence interval of 2 times 0.9975=0.9950, that is, the width of the area below the distribution curve. We can be 99.5% confident that our data will be 2.81 standard deviations below or above the mean. </p>
<p style="text-align:justify">A 2-tailed distribution means that we are interested in the distribution on either side of the mean, and not just on one side or the other (only greater than or only less than).</p>Statistics Answershttps://www.mathhomeworkanswers.org/261405/find-the-critical-value-za-corresponding-confidence-level?show=261406#a261406Fri, 17 May 2019 20:56:52 +0000Answered: 150 randomly selected people were surveyed about the location of the concession stand.
https://www.mathhomeworkanswers.org/261398/randomly-selected-people-surveyed-about-location-concession?show=261404#a261404
<p style="text-align:justify">The proportion of people who felt the stand should be closer to the bleachers is 108/150=18/25=0.72.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261398/randomly-selected-people-surveyed-about-location-concession?show=261404#a261404Fri, 17 May 2019 20:18:08 +0000Answered: use a sample n = 500, p^ = 0.90, and a 95% confidence level to construct a confidence interval estimate of the population proportion p: Answer 0.874 < p < 0.926 How do you get that answer?
https://www.mathhomeworkanswers.org/261392/confidence-construct-confidence-estimate-population-proportion?show=261395#a261395
<p style="text-align:justify">This looks like a binomial distribution where mean=np and variance=np(1-p). In this question p=sample proportion p hat.</p>
<p style="text-align:justify">So, with mean=500×0.9=450, variance=500×0.9×0.1=45, standard deviation=√45=6.708 approx.</p>
<p style="text-align:justify">The standard error is standard deviation/n=6.708/500=0.0134 approx (I think!).</p>
<p style="text-align:justify">Now we need a t test critical value for 95% confidence level. The large sample size means we can use 1.96 as the critical value, which we get from t tables or normal distribution (z) tables for a significance level of 0.05.</p>
<p style="text-align:justify">We apply this critical value to the standard error to get the margin of error for the required confidence interval=1.96×0.0134=0.026 approx.</p>
<p style="text-align:justify">The test statistic is p=0.90, and the 95% confidence level means that we can apply a range of values for p: 0.90±0.026.</p>
<p style="text-align:justify">Therefore 0.900-0.026<p<0.900+0.026, that is, 0.874<p<0.926.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261392/confidence-construct-confidence-estimate-population-proportion?show=261395#a261395Fri, 17 May 2019 16:41:05 +0000Answered: Statistics question below in more information
https://www.mathhomeworkanswers.org/261206/statistics-question-below-in-more-information?show=261219#a261219
<p style="text-align:justify">The probability is the sum of the individual probabilities less the probability that H and W both watch the TV show=0.84-0.11=0.73.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261206/statistics-question-below-in-more-information?show=261219#a261219Mon, 13 May 2019 17:10:20 +0000Answered: Sketch the region corresponding to the statement P ( − 2.2 < z < − 0.7 )
https://www.mathhomeworkanswers.org/261192/sketch-the-region-corresponding-to-the-statement-p-%E2%88%92-2-2-z-%E2%88%92-0-7?show=261214#a261214
<p style="text-align:justify"><img alt="" src="https://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=4679009705977977397" style="height:93px; width:450px"></p>
<p style="text-align:justify">The open interval (-2.2,-0.7) is shown on a number line. The variable z can be anywhere on the red line excluding the endpoints.</p>Statistics Answershttps://www.mathhomeworkanswers.org/261192/sketch-the-region-corresponding-to-the-statement-p-%E2%88%92-2-2-z-%E2%88%92-0-7?show=261214#a261214Mon, 13 May 2019 16:48:14 +0000Answered: what is the the range in this data, the variance, and the standard deviation
https://www.mathhomeworkanswers.org/75757/what-the-the-range-this-data-the-variance-standard-deviation?show=261123#a261123
<p style="text-align:justify">The range is the difference between the highest and lowest data: 385-0=385.</p>
<p style="text-align:justify">To find the variance first find the average (mean) of all the data. Add the data values together and then divide by the amount of data (15 in this case): 3293/15=219.53.</p>
<p style="text-align:justify">We subtract the mean from each datum in the set and square the result. Then we add together all of these and divide by 15: 10711.58 (approx). Another way to do this is to add up all the squares of the data, divide the result by 15, then subtract the square of the mean. </p>
<p style="text-align:justify">We take the square root of this variance to get the standard deviation: 103.50 (approx).</p>Statistics Answershttps://www.mathhomeworkanswers.org/75757/what-the-the-range-this-data-the-variance-standard-deviation?show=261123#a261123Fri, 10 May 2019 09:24:31 +0000Answered: How many 5 character license plates can I create if the first 2 characters must be letters and the last 3 must be numbers?
https://www.mathhomeworkanswers.org/261101/character-license-plates-create-characters-letters-numbers?show=261104#a261104
Number of letters in English =26<br />
<br />
The numbers can be 000 to 999 ie 1000 in total <br />
<br />
or (0-9) ie 10 for each digit so 10^3 =1000 for 3 digits<br />
<br />
The number of 5 character license plates =(26)(26)(1000)=676000Statistics Answershttps://www.mathhomeworkanswers.org/261101/character-license-plates-create-characters-letters-numbers?show=261104#a261104Fri, 10 May 2019 02:03:54 +0000