9.11 Hypothesis testing of single mean and single proportion:  (Page 14/4)

Set up the Hypothesis Test:

Since the problem is about a mean weight, this is a test of a single population mean .

$H_{\mathrm{o}}$ : $\mu$ $=275$ $H_a$ : $\mu$ $>275$ This is a right-tailed test.

Calculating the distribution needed:

Random variable: $\overline{X}$ = the mean weight, in pounds, lifted by the football players.

Distribution for the test: It is student t-test because $\sigma$ is unknown, so we will need to check normal distribution assumptions by looking at a histogram and box plot of the data as well as the descriptive statistics. The data is slightly skewed to the right with no outliers, so we will proceed with some reservation since the data is varies from a normal distribution. The median is 8 points higher than the mean. The sample standard deviation is 55.9. We will round the standard deviation to 56 for calculating the p-value.

$\overline{X}$ ~ $N$ $(275,\frac{56}{\sqrt{30}})$

$\overline{x}$ $=286.2$ pounds (from the data).

$s=56$ pounds We assume $\mu =275$ pounds unless our data shows us otherwise.

Calculate the p-value using the normal distribution for a mean and using the sample mean as input:

$\text{p-value}=P($ $\overline{x}$ $>286.2$ ) $=0.1414$ .

Interpretation of the p-value: If $H_{\mathrm{o}}$ is true, then there is a 0.1414 probability (14.14%) that the football players can lift a mean weight of 286.2pounds or more. Because a 14.14% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.

Compare $\alpha$ and the p-value:

$\alpha =0.025\text{p-value}=0.1414$

Make a decision: Since $\alpha$ $$ $\text{p-value}$ , do not reject $H_{\mathrm{o}}$ .

Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275pounds.