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Leokris [45]
3 years ago
8

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.
Mathematics
1 answer:
Oksanka [162]3 years ago
5 0

Answer: -3.0593

Step-by-step explanation:

The test statistic for the difference between two population mean (when population standard deviations \sigma_1\ \&\ \sigma_2  are known ) is given by :-

z=\dfrac{\overline{x}_1-\overline{x}_2}{\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}}

where ,

n_1\ \&\ n_2 = Sample sizes taken from two  populations 1 and 2.

\overline{x}_1-\overline{x}_2 = Difference between two Sample mean.

\sigma_1\ \&\ \sigma_2 = Standard deviations of populations 1 and 2.

As per given , we have

n_1=55\ \&\ n_2=58

\overline{x}_1=2.95\ \&\ \overline{x}_2=3.3

\sigma_1=0.65\ \&\ \sigma_2=0.56

We assume that the data follows a normal distribution.

Then, the test statistic will be :-

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  

Hence, the test statistic= -3.0593

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