Question

The average American gets a haircut every 43 days. Is the average larger for college students?...

The average American gets a haircut every 43 days. Is the average larger for college students? The data below shows the results of a survey of 12 college students asking them how many days elapse between haircuts. Assume that the distribution of the population is normal.

45, 53, 39, 46, 37, 47, 43, 40, 35, 53, 39, 55

What can be concluded at the the α = 0.10 level of significance level of significance?

A) For this study, we should use (Z-Test for population proportion or T-Test for population mean)

B) The null and alternative hypotheses would be:

H 0 : (?,P,μ) (?,<,>,=,≠) Answer:__________

H 1 : (?,P,μ) (?,<,>,=,≠) Answer:__________

C) The test statistic = (?,t,z) = _________ (please show your answer to 3 decimal places.)

D) The p-value = __________(Please show your answer to 3 decimal places.)

E) The p-value is α (?, ≥,≤ )

F) Based on this, we should the null hypothesis. (Reject,Fail to Reject, or Accept)

G) Thus, the final conclusion is that ...

Mark an X on the correct answer below

___The data suggest the population mean is not significantly higher than 43 at α = 0.10, so there is sufficient evidence to conclude that the population mean number of days between haircuts for college students is equal to 43.

___The data suggest the population mean number of days between haircuts for college students is not significantly higher than 43 at α = 0.10, so there is insufficient evidence to conclude that the population mean number of days between haircuts for college students is higher than 43.

___The data suggest the populaton mean is significantly higher than 43 at α = 0.10, so there is sufficient evidence to conclude that the population mean number of days between haircuts for college students is higher than 43.

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Answer #1

a)

T-Test for population mean

b)

Null Hypothesis: H0 : μ = 43
Alternative Hypothesis: H1: μ > 43

c)

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (44.1667 - 43)/(7.0156/sqrt(10))
t = 0.526

d)

P-value = 0.3058

e)


As P-value >= 0.1,

f)

fail to reject null hypothesis.


g)

The data suggest the population mean number of days between haircuts for college students is not significantly higher than 43 at α = 0.10, so there is insufficient evidence to conclude that the population mean number of days between haircuts for college students is higher than 43.

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