an adventure company runs two obstacle courses, fundash and coolsprint, with similar designs. since fundash…

an adventure company runs two obstacle courses, fundash and coolsprint, with similar designs. since fundash was built on rougher terrain, the designer of the courses suspects that the mean completion time of fundash is greater than the mean completion time of coolsprint. to test this, she selects 235 fundash runners and 265 coolsprint runners. (consider these as independent random samples of the fundash and coolspring runners.) the 235 fundash runners complete the course with a mean time of 77.9 minutes and a standard deviation of 8.6 minutes. the 265 individuals complete coolsprint with a mean time of 75.7 minutes and a standard deviation of 7.9 minutes. assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. at the 0.01 level of significance, is there enough evidence to support the claim that the mean completion time, μ1, of fundash is greater than the mean completion time, μ2, of coolsprint? perform a one - tailed test. then complete the parts below. carry your intermediate computations to at least three decimal places. (if necessary, consult a list of formulas.) (a) state the null hypothesis h0 and the alternative hypothesis h1. h0:□ h1:□ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) □ (d) find the critical value at the 0.01 level of significance. (round to three or more decimal places.) □ (e) can we support the claim that the mean completion time of fundash is greater than the mean completion time of coolsprint? yes no

an adventure company runs two obstacle courses, fundash and coolsprint, with similar designs. since fundash was built on rougher terrain, the designer of the courses suspects that the mean completion time of fundash is greater than the mean completion time of coolsprint. to test this, she selects 235 fundash runners and 265 coolsprint runners. (consider these as independent random samples of the fundash and coolspring runners.) the 235 fundash runners complete the course with a mean time of 77.9 minutes and a standard deviation of 8.6 minutes. the 265 individuals complete coolsprint with a mean time of 75.7 minutes and a standard deviation of 7.9 minutes. assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. at the 0.01 level of significance, is there enough evidence to support the claim that the mean completion time, μ1, of fundash is greater than the mean completion time, μ2, of coolsprint? perform a one - tailed test. then complete the parts below. carry your intermediate computations to at least three decimal places. (if necessary, consult a list of formulas.) (a) state the null hypothesis h0 and the alternative hypothesis h1. h0:□ h1:□ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) □ (d) find the critical value at the 0.01 level of significance. (round to three or more decimal places.) □ (e) can we support the claim that the mean completion time of fundash is greater than the mean completion time of coolsprint? yes no

Answer

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the mean completion - time of Fundash is less than or equal to the mean completion - time of Coolsprint, i.e., $H_0:\mu_1\leq\mu_2$ or $H_0:\mu_1 - \mu_2\leq0$. The alternative hypothesis $H_1$ is that the mean completion - time of Fundash is greater than the mean completion - time of Coolsprint, i.e., $H_1:\mu_1>\mu_2$ or $H_1:\mu_1 - \mu_2>0$.

Step2: Determine test statistic type

Since the population standard deviations are estimated from large samples, we use a two - sample z - test.

Step3: Calculate the test statistic

The formula for the two - sample z - test statistic is $z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}$. Here, $\bar{x}_1 = 77.9$, $s_1 = 8.6$, $n_1 = 235$, $\bar{x}_2 = 75.7$, $s_2 = 7.9$, $n_2 = 265$, and under $H_0$, $\mu_1-\mu_2 = 0$. [ \begin{align*} z&=\frac{(77.9 - 75.7)-0}{\sqrt{\frac{8.6^{2}}{235}+\frac{7.9^{2}}{265}}}\ &=\frac{2.2}{\sqrt{\frac{73.96}{235}+\frac{62.41}{265}}}\ &=\frac{2.2}{\sqrt{0.314723 + 0.23551}\ &=\frac{2.2}{\sqrt{0.550233}}\ &=\frac{2.2}{0.741789}\ &\approx2.966 \end{align*} ]

Step4: Find the critical value

For a one - tailed test with a significance level of $\alpha = 0.01$, the critical value $z_{\alpha}$ is such that $P(Z>z_{\alpha})=0.01$. Looking up in the standard normal table, $z_{0.01}\approx2.326$.

Step5: Make a decision

Since the calculated test statistic $z = 2.966>z_{0.01}=2.326$, we reject the null hypothesis.

Answer:

(a) $H_0:\mu_1 - \mu_2\leq0$, $H_1:\mu_1 - \mu_2>0$ (b) Two - sample z - test (c) $2.966$ (d) $2.326$ (e) Yes