A diet plan claims that women have a mean weight of 145 lbs. Assume that the weights of women have a standard deviation of 30.86 lb. A sample of 40 weights of women has a mean weight of 153.2 lb. Find the p-value and, using a 0.05 significance level, state the conclusion about the null hypothesis. Could someone help me solve this?

SAMPLE STATISTICS: mean (x bar)=153.2lb, size (n)=40 women.

POPULATION STATISTIC: standard deviation (σ)=30.86lb, mean (µ)=145lb.

CLAIM: Women have a mean weight of 145lbs.

COUNTERCLAIM: The mean weight of women is not equal to 145lbs.

Null hypothesis, H₀: µ=145lbs

Alternative hypothesis, H₁: µ≠145lbs

SIGNIFICANCE LEVEL: ɑ=0.05 (95% confidence level), 2-tail.

TEST STATISTIC: Z=(x bar-µ)/(σ/√n)=(153.2-145)/(30.86/√40),

Z=8.2/4.88=1.68.

To find p-value, look up Z=-1.68 or look up Z=1.68 and subtract from 1.

p-value=0.0465.

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.

(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.)

by Top Rated User (680k points)