the sical demands of women\'s vens is rapidly growing in popularity g ded in the
ID: 3362960 • Letter: T
Question
the sical demands of women's vens is rapidly growing in popularity g ded in the 2016 Olympics. Matches are 8 Physical 58 will be include and l on a full and co played halves. Each team also consists of se merstand the demands of women's gby field and consist of two seven- ven pla minute better u oup of researchers compared the physical To sevens, a SeVies of elite players from the Canadian National dm with a university squad. The following ta ummarizes some of these qualities:28 tabe Elite (n=16) University (n-13) Quality Sprint speed (km/hr) Peak heart rate (bpm) Intermittent recovery test (m) 27.3 192.0 0.7 6.0 191 26.0 193.0 781 6.0 129 carry out the significance tests using -0.05. Report the test statistic with the degrees of freedom and the P-value. Write a short summary of your conclusion.Explanation / Answer
Solution:-
1) Test for sprint speed.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
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 = 0.4513
DF = 27
t = [ (x1 - x2) - d ] / SE
t = 2.88
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.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 27 degrees of freedom is more extreme than -2.88 that is, less than - 2.88 or greater than 2.88.
Thus, the P-value = 0.0038
Interpret results. Since the P-value (0.0038) is less than the significance level (0.05), we cannot accept the null hypothesis.
ii) Test for peak heart rate
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
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 = 2.24
DF = 27
t = [ (x1 - x2) - d ] / SE
t = - 0.45
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.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 27 degrees of freedom is more extreme than -0.45 that is, less than - 45 or greater than 0.45
Thus, the P-value = 0.6564
Interpret results. Since the P-value (0.6564) is greater than the significance level (0.05), we have to accept the null hypothesis.
iii) Test for intermittent recovery
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
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 = 59.67
DF = 27
t = [ (x1 - x2) - d ] / SE
t = 6.4
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.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 27 degrees of freedom is more extreme than 6.4 that is, less than - 6.4 or greater than 6.4
Thus, the P-value = less than 0.00001
Interpret results. Since the P-value (almsot 0) is less than the significance level (0.05), we cannot accept the null hypothesis.
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