Q:

Among a sample of 55 male soldiers the average shower time was found to be 2.95 minutes and the population standard deviation is known to be 0.65 minutes. Among a sample of 58 female soldiers the average shower time was found to be 3.3 minutes and the population standard deviation is known to be 0.56 minutes. Assume normality of the data. What is the test statistic? Give your answer to four decimal places.

Accepted Solution

A:
Answer: -3.0593Step-by-step explanation:The test statistic for the difference between two population mean (when population standard deviations [tex]\sigma_1\ \&\ \sigma_2[/tex]  are known ) is given by :-[tex]z=\dfrac{\overline{x}_1-\overline{x}_2}{\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}}[/tex]where , [tex]n_1\ \&\ n_2[/tex] = Sample sizes taken from two  populations 1 and 2.[tex]\overline{x}_1-\overline{x}_2[/tex] = Difference between two Sample mean.[tex]\sigma_1\ \&\ \sigma_2[/tex] = Standard deviations of populations 1 and 2.As per given , we have[tex]n_1=55\ \&\ n_2=58[/tex] [tex]\overline{x}_1=2.95\ \&\ \overline{x}_2=3.3[/tex] [tex]\sigma_1=0.65\ \&\ \sigma_2=0.56[/tex] We assume that the data follows a normal distribution.Then, the test statistic will be :-[tex]z=\dfrac{2.95-3.3}{\sqrt{\dfrac{(0.65)^2}{55}+\dfrac{(0.56)^2}{58}}}\\\\=\dfrac{-0.35}{\sqrt{0.0130887}}\\\\=\dfrac{-0.35}{0.1144059}=3.05928278174\approx-3.0593[/tex]  Hence, the test statistic= -3.0593