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

Answer:

0.0930; fail to reject the null hypothesis

Step-by-step 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

$H_1:\mu\neq145$

Significance Level

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

Test Statistic

$Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{153.2-145}{\frac{30.86}{\sqrt{40}}}\approx1.6805$

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.