Question

A company is developing a new high performance wax for cross country ski racing. In order to justify the price marketing​ wants, the wax needs to be very fast.​ Specifically, the mean time to finish their standard test course should be less than 5555 seconds for a former Olympic champion. To test​ it, the champion will ski the course 8 times. The​ champion's times​ (selected at​ random) are 50.450.4​, 65.965.9​, 49.549.5​, 50.950.9​, 48.248.2​, 48.548.5​, 53.853.8​, and 43.543.5 seconds to complete the test course. Should they market the​ wax? Assume the assumptions and conditions for appropriate hypothesis testing are met for the sample. Use 0.05 as the​ P-value cutoff level.

Choose the correct null and alternative hypotheses below. O A. Ho:p=55 HA:u> 55 OC. He:p=55 HA:< 55 OB. Ho:<55 Hair= 55 OD. Ho:p> 55 Hair = 55

Calculate the test statistic.

Calculate the​ P-value.

Answer 1

The null and alternative hypothesis for the test is:

$\mathrm{H_{0}:\mu=55}$; i..e, the true mean time to finish the standard test course is not different than 55 seconds.

$\mathrm{H_{A}:\mu<55}$; i.e., the true mean time to finish the standard test course is less than 55 seconds.

Test-statistic: $\mathrm{t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}}$ and it follows a t-distribution with degrees of freedom, $\mathrm{df=n-1}$

sample mean, $\bar{x}:$

$\mathrm{\bar{x}=\frac{50.4&plus;65.9&plus;49.5&plus;50.9&plus;48.2&plus;48.5&plus;53.8&plus;43.5}{8}=\frac{410.7}{8}=\mathbf{51.3375}}$

sample standard deviation, $s:$

$\mathrm{s=\sqrt{\frac{\Sigma{(x-\bar{x})^{2}}}{n-1}}=\sqrt{\frac{301.8988}{8-1}}=\mathbf{6.567222}}$

$\mathrm{\mathrm{t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{51.3375-55}{\frac{6.567222}{\sqrt{8}}}=-1.577396705\approx \mathbf{-1.577}}}$

So, the test-statistic is calculated as $\mathbf{t=-1.577}$

P-value: Since it is a left-tailed test, so the p-value for the test-statistic is calculated as-

$\mathrm{p-value=P(t<-1.577,\:df=8-1=7)=0.07940066\approx \mathbf{0.0794}}$

So, the p-value for the hypothesis test is calculated as $\mathbf{p-value=0.0794}$

Decision: $\mathrm{\alpha=0.05,\:\:p-value=0.0794}$

Since, $\mathrm{p-value>\alpha}\Rightarrow \mathbf{We\:fail\:to\:reject\:null\:hypothesis\:H_{0}}$

So at 5% significance level the sample data does not provides sufficient evidence to support the alternative hypothesis, hence we fail to reject the null hypothesis H_0.