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A facilities manager at a university reads in a research report that the mean amount of time spent in the shower by an adult is 5 minutes. He decides to collect data to see if the mean amount of time that college students spend in the shower is significantly different from 5 minutes. In a sample of 15 students, he found the average time was 4.29 minutes and the standard deviation was 0.75 minutes. Using this sample information, conduct the appropriate hypothesis test at the 0.01 level of significance. Assume normality.

Sagot :

Ogorwyneidn

Step-by-step explanation:

n = 15

average = bar x = 4.29

null hypothesis

h0: μ=5

alternative hypothesis=

h1 = μ≠5

we find the test statistics

t stat = [tex]\frac{4.29-5}{0.75/\sqrt{15} }[/tex]

= -0.71/0.1936

= -3.6673

degree of freedom = 15 - 1 = 14

we find p value using excel's t distribution function

= T.DIST(3.664,14 2) this is a 2 tailed test

p value = 0.0025

alpha = 0.01

because 0.0025 < 0.01

conclusion

so we reject the claim that the time = 5 minutes  since p value is smaller than alpha.

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