Question

An engineer in charge of water rationing for the U.S. Army wants to determine if the average male soldier spends less time in the shower than the average female soldier. Let μm represent the average time in the shower of male soldiers and μf represent the average time in the shower of female soldiers.
a) What are the appropriate hypotheses for the engineer?
H0: μm = μf versus Ha: μm > μf
H0: σm = σf versus Ha: σm > σf
H0: μm = μf versus Ha: μm ≠ μf
H0: μm = μf versus Ha: μm < μf

b) Among a sample of 66 male soldiers the average shower time was found to be 2.68 minutes and the standard deviation was found to be 0.65 minutes. Among a sample of 69 female soldiers the average shower time was found to be 2.7 minutes and the standard deviation was found to be 0.5 minutes. What is the test statistic? Give your answer to three decimal places.
c) What is the P-value for the test? Give your answer to four decimal places.
d) Using a 0.1 level of significance, what is the appropriate conclusion?
Reject the claim that the average shower times are different for male and female soldiers because the P-value is greater than 0.1.
Conclude that the average shower time for males is less than the average shower time for females because the P-value is less than 0.1.
Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.
Conclude that the average shower time for males is equal to the average shower time for females because the P-value is less than 0.1.

Answer 1

Answer:

a) H0: μm = μf versus Ha: μm < μf

b) $t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200$

c) $p_v =P(t_{133}$  

d) Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.

Step-by-step explanation:

Data given and notation  

$\bar X_{m}=2.68$ represent the mean for the sample male

$\bar X_{f}=2.7$ represent the mean for the sample female

$s_{m}=0.65$ represent the sample standard deviation for the males

$s_{f}=0.5$ represent the sample standard deviation for the females  

$n_{m}=66$ sample size for the group male  

$n_{f}=69$ sample size for the group female  

t would represent the statistic (variable of interest)  

Part a

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the average male soldier spends less time in the shower than the average female soldier, the system of hypothesis would be:  

Null hypothesis:$\mu_{m}-\mu_{f}\geq 0$  

Alternative hypothesis:$\mu_{m} - \mu_{f}< 0$  

Or equivalently:

Null hypothesis:$\mu_{m}-\mu_{f}= 0$  

Alternative hypothesis:$\mu_{m} - \mu_{f}< 0$  

And the best option is:

H0: μm = μf versus Ha: μm < μf

Part b

We don't have the population standard deviation, so for this case is better apply a t test to compare means, and the statistic is given by:  

$t=\frac{(\bar X_{m}-\bar X_{f})-\Delta}{\sqrt{\frac{s^2_{m}}{n_{m}}+\frac{s^2_{f}}{n_{f}}}}$ (1)

And the degrees of freedom are given by $df=n_m +n_f -2=66+69-2=133$  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

What is the test statistic?

With the info given we can replace in formula (1) like this:  

$t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200$

Part c What is the p-value?

Since is a left tailed test the p value would be:  

$p_v =P(t_{133}$  

Part d

The significance level given is $\alpha =0.1$ since the p value is higher than the significance level we can conclude:

Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.