Question

A researcher wished to compare the effect of two stepping heights (low and high) on heart rate in a step-aerobics workout. A collection of 50 adult volunteers was randomly divided into two groups of 25 subjects each. Group 1 did a standard step-aerobics workout at the low height. The mean heart rate at the end of the workout for the subjects in Group 1 was = 90.00 beats per minute with a standard deviation of = 9 beats per minute. Group 2 did the same workout but at the high step height. The mean heart rate at the end of the workout for the subjects in Group 2 was = 95.08 beats per minute with a standard deviation of = 12 beats per minute. Assume that two groups are independent and the data are approximately Normal. Let and represent the mean heart rates we would observe for the entire population represented by the volunteers, if all members of this population did the workout using the low or high step height, respectively. Suppose the researcher had wished to test the hypotheses and . The P-value for the test is (use conservative value for the degrees of freedom)

Answer 1

Answer:

The answer is "between 0.10 and 0.05".

Step-by-step explanation:

For sample 1:

$n_1 = 25\\\\\bar{x_1} = 90.00\\\\ s_1 = 9$

For sample 2:

$n_2 = 25\\\\ \bar{x_2} = 95.08\\\\ S_2 = 12$

Calculating the null and alternative hypotheses:

$H_O : \mu_1 = \mu_2 \\\\ H_Q : \mu_1$  

Calculating the test statistic:

$Z= \frac{90-95.08}{ \sqrt{\frac{9^2}{25}+\frac{12^2}{25}}} \\\\ =\frac{-5.08}{ \sqrt{\frac{81}{25}+\frac{144}{25}}}\\\\=\frac{-5.08}{ \sqrt{\frac{81+144}{25}}}\\\\=\frac{-5.08}{ \sqrt{\frac{225}{25}}}\\\\=\frac{-5.08}{ \frac{15}{5}}\\\\=\frac{-5.08}{3}\\\\= -1.693$

Calculating the conservative degrees of freedom:

$DF = min (n_1 -1, n_2 - 1) = min(25-1, 25-1) = 24$

by using Excel the p-value for left tailed test and for test statistic will be $-1.693$ with $DF = 24$.

$\to p-value = TDIST(1.693, 24, 1) = 0.0517\ \ \ \ i.e, \bold{0.05 < p-value < 0.10}$