section 8.3 testing the difference between means (dependent samples) 443\n11. caffeine ingestion a…

section 8.3 testing the difference between means (dependent samples) 443\n11. caffeine ingestion a researcher claims that caffeine ingestion improves repeated freestyle sprints in trained male swimmers. the table shows the mean performance times (in seconds) for a group of trained male swimmers who complete six 75 - meter maximal freestyle sprints after ingesting a placebo and after ingesting caffeine. at $alpha = 0.01$, is there enough evidence to support the researchers claim? (source: journal of sports science & medicine)\nsprint number 1 2 3 4 5 6\nsprint time (with placebo) 40.2 40.3 40.7 41.0 40.7 40.6\nsprint time (with caffeine) 39.9 39.9 39.7 40.1 40.2 40.5

section 8.3 testing the difference between means (dependent samples) 443\n11. caffeine ingestion a researcher claims that caffeine ingestion improves repeated freestyle sprints in trained male swimmers. the table shows the mean performance times (in seconds) for a group of trained male swimmers who complete six 75 - meter maximal freestyle sprints after ingesting a placebo and after ingesting caffeine. at $alpha = 0.01$, is there enough evidence to support the researchers claim? (source: journal of sports science & medicine)\nsprint number 1 2 3 4 5 6\nsprint time (with placebo) 40.2 40.3 40.7 41.0 40.7 40.6\nsprint time (with caffeine) 39.9 39.9 39.7 40.1 40.2 40.5

Answer

Explanation:

Step1: Calculate the differences

Let $d_i$ be the difference between the sprint - time with placebo and the sprint - time with caffeine for each sprint. $d_1=40.2 - 39.9=0.3$, $d_2=40.3 - 39.9 = 0.4$, $d_3=40.7-39.7 = 1.0$, $d_4=41.0 - 40.1=0.9$, $d_5=40.7 - 40.2 = 0.5$, $d_6=40.6 - 40.5=0.1$.

Step2: Calculate the mean of the differences $\bar{d}$

$\bar{d}=\frac{\sum_{i = 1}^{6}d_i}{n}$, where $n = 6$ and $\sum_{i=1}^{6}d_i=0.3 + 0.4+1.0 + 0.9+0.5+0.1=3.2$. So, $\bar{d}=\frac{3.2}{6}\approx0.533$.

Step3: Calculate the standard deviation of the differences $s_d$

First, calculate $(d_i-\bar{d})^2$ for each $i$: $(d_1 - \bar{d})^2=(0.3 - 0.533)^2=(- 0.233)^2 = 0.0543$, $(d_2-\bar{d})^2=(0.4 - 0.533)^2=(-0.133)^2 = 0.0177$, $(d_3-\bar{d})^2=(1.0 - 0.533)^2=(0.467)^2 = 0.2181$, $(d_4-\bar{d})^2=(0.9 - 0.533)^2=(0.367)^2 = 0.1347$, $(d_5-\bar{d})^2=(0.5 - 0.533)^2=(-0.033)^2 = 0.0011$, $(d_6-\bar{d})^2=(0.1 - 0.533)^2=(-0.433)^2 = 0.1875$. $\sum_{i = 1}^{6}(d_i-\bar{d})^2=0.0543+0.0177 + 0.2181+0.1347+0.0011+0.1875 = 0.6134$. $s_d=\sqrt{\frac{\sum_{i = 1}^{6}(d_i-\bar{d})^2}{n - 1}}=\sqrt{\frac{0.6134}{5}}\approx0.35$.

Step4: Calculate the test - statistic $t$

The null hypothesis $H_0:\mu_d\leq0$ and the alternative hypothesis $H_1:\mu_d>0$. The test - statistic $t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}}$, with $\mu_d = 0$ (under $H_0$), $n = 6$, $\bar{d}\approx0.533$, and $s_d\approx0.35$. $t=\frac{0.533-0}{0.35/\sqrt{6}}\approx3.73$.

Step5: Determine the critical value

The degrees of freedom $df=n - 1=6 - 1 = 5$. For a one - tailed test with $\alpha=0.01$ and $df = 5$, the critical value $t_{\alpha,df}=t_{0.01,5}=3.365$.

Step6: Make a decision

Since the calculated $t\approx3.73>t_{0.01,5}=3.365$, we reject the null hypothesis.

Answer:

Yes, there is enough evidence at the $\alpha = 0.01$ level of significance to support the researcher's claim that caffeine ingestion improves repeated freestyle sprints in trained male swimmers.