Question

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. 0.0465; reject the null hypothesis 0.0930; fail to reject the null hypothesis

Answer 1

0.0930; fail to reject the null hypothesis

Explanation:

Sample Statistics

`\bar{x}=153.2`

`n=40`

Population Statistics

`\sigma=30.86`

`\mu=145`

Claim

The mean weight of women is 145lbs

Counterclaim:

The mean weight of women is not equal to 145lbs

Null hypothesis

`H_0:\mu=145`

Alternative hypothesis

`H_1:\mu\\eq145`

Significance Level

`\alpha=0.05` for a 2-tailed test at a 95% confidence level

Test Statistic

`Z=\frac{\bar{x}-\mu}{(\sigma)/(√(n))}=(153.2-145)/((30.86)/(√(40)))\approx1.6805`

Corresponding p-value

`p=2[$normalcdf(1.6805,\infty,0,1)]\approx2(0.04643)\approx0.09286\approx0.0930`

The test is 2-tailed because our alternate hypothesis,
`H_1`, already implies that the population mean,
`\mu`, could be either less than or greater than 145lb. Since 0.09286>0.05, we are within the 95% confidence level, so there's insufficient evidence in our sample data to suggest that we reject the null hypothesis. This means that it's 9.3% more likely that the null hypothesis is true than the alternate hypothesis is. Thus, the sample data do not differ significantly from the expected mean of 145lb.