The Cougar Swim Club acquired some Speedo Fastskin bodysuits and decided to test
ID: 3259162 • Letter: T
Question
The Cougar Swim Club acquired some Speedo Fastskin bodysuits and decided to test them out. A number of the club's fastest swimmers performed a 50m freestyle swim in a regular spandex bodysuit and in a Speedo Fastskin suit. The table below summarizes their times in seconds (s). Perform a t-test for dependent means to determine if there is a difference between the regular spandex suit and the Fastskin bodysuit in terms of performance. t = df = Critical value of t = (use alpha = 0.05) Would you reject the null hypothesis?Explanation / Answer
Solution:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0 , there is no difference between the regular spandex suit and the Fastskin bodysuit in terms of performance.
Alternative hypothesis: 1 - 2 0 , there is difference between the regular spandex suit and the Fastskin bodysuit in terms of performance.
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(2.992/8) + (2.5852/8)] = 1.397
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (2.992/8 + 2.5852/8)2 / { [ (2.992 / 8)2 / (7) ] + [ (2.5852 / 8)2 / (7) ] }
DF = 3.81339 / (0.1784 + 0.0996699) = 13.71 or `14
t = [ (x1 - x2) - d ] / SE = [ (31.85 - 29.05) - 0 ] / 1.397 = 2.0042949 or 2.00
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Crtical value of t = + 2.14478669 or + 2.14
We use the t Distribution Calculator to find P(t < 2.00)
The P-Value is 0.06479.
The result is not significant at p < 0.05.
Interpret results. Since the P-value (0.065) is greater than the significance level (0.05), we cannot reject the null hypothesis.
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