statistics question on hypothesis and probability

Question

1. Define Hypothesis testing and explain briefly the five standard steps involved in this process.

2. The long-distance calls made by the employees of a company are normally distributed with a mean of 6.3 minutes and a standard deviation of 2.2 minutes.
Find the probability that a call:
2.1. Lasts between 5 and 10 minutes.
2.2. Lasts more than 7 minutes.
2.3. Lasts less than 4 minutes

 

3. A manufacturer of light bulbs advertises that, on average, its long-life bulb will last more than 5 000 hours. To test the claim, a statistician took a random sample of

100 bulbs and measured the amount of time each bulb burned out. She found that the sample mean was 5065 hours. If we assume that the lifetime of this type of bulb is normally distributed and has a standard deviation of 400 hours, test at 5% level of
significance the manufacture’s claim is true.

Answer 1

1. The null hypothesis, H₀, is the prevalent status quo value for the mean or proportion, while the alternative hypothesis, H_a is a claim that the value is something different: less than, greater than or simply not equal to this value. Testing involves taking the difference of the claimed mean and the prevalent mean and working out how many standard deviations this difference is from the prevalent mean. If this difference exceeds a particular critical value the null hypothesis is rejected and the alternative hypothesis is more likely. Otherwise, the null hypothesis cannot be rejected and is therefore probably still true (countering the alternative hypothesis).

The critical value depends on three things:

• the sample size of the evidence data involved in the claim (this number less 1 gives the number of degrees of freedom when looking up the critical value)
• whether the claim is for just a different mean (either greater than or less than); or specifically greater or less than the mean, 2-tailed test or 1-tailed test respectively
• the significance or confidence level (specified degree of certainty)

Tables or a calculator give the required critical value.

2. The Z-score, Z=(X-μ)/σ, that is, the number of standard deviations from the mean of some data value X. Tables or a calculator convert this score into a probability. 

2.1 Z₁=(5-6.3)/2.2=-0.59, Z₂=(10-6.3)/2.2=1.68 approx. These correspond to P₁=0.277, P₂=0.954. P₂-P₁=0.677. So P(5<X<10)=67.7% approx.

2.2 Z=(7-6.3)/2.2=1-0.625=0.375, or 37.5% approx. Tables give P(X<7)=0.625, so P(X>7)=1-0.625=0.375.

2.3 Z=(4-6.3)/2.2=-1.045. P(X<4)=0.148, or 14.8% approx.

3. Null hypothesis, H₀, μ=5000. Alternative hypothesis, H_a, μ>5000. Sample size=100 so degrees of freedom=99. 

Z=(5065-5000)/400=0.1625. For this sample size (100, quite large) critical value (1-tailed) corresponding to 5% significance (95% confidence level) is 1.98. 0.1625<1.98, therefore H₀ cannot be rejected, which means 5000 hours is still deemed to be the average bulb life-time, and the manufacturer's claim cannot be disputed. We can be 95% certain that this is the case.