Question

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The average wait time to see a doctor at the Walnut ER on a Monday afternoon is 12 minutes. Is the mean wait time less for women with small children? Wait times for 12 randomly selected women with small children are shown below. Assume that the distribution of the population is normal. 11, 11, 9, 12, 13, 11, 10, 10, 11, 12, 12, 11 What can be concluded at the the a -0.01 level of significance level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: HO: ? Select an answer H: ? Select an answer C. The test statistic ? - (please round your answer to 3 decimal places.) d. The p-value (Please round your answer to 4 decimal places.) e. The p-value is a f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... The data suggest the populaton mean is significantly less than 12 at a -0.01, so there is statistically significant evidence to conclude that the population mean wait time for women with small children is less than 12. The data suggest the population mean is not significantly less than 12 at a = 0.01, so there is statistically insignificant evidence to conclude that the population mean wait time for women with small children is equal to 12. The data suggest that the population mean wait time for women with small children is not significantly less than 12 at a = 0.01, so there is statistically insignificant evidence to conclude that the population mean wait time for women with small children is less than 12.

Answer 1

(Note to student : Since you have not provided options to the blanks, I am completing the statement with logical answers. Use the options closest to the answers.)

	n	Mean (M)	Variance (SS)	Standard Deviation (SD)
Wait Times	n = 12	X̅ = 11.0833	Var = 1.1742	s = 1.0836

α = 0.01

We use single sample T test since population standard deviation is unknown, and sample size is small

a) For this study we should use t test

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b)

The null and alternative hypotheses would be
Ho :  μ = 12                      ( or μ ≥ 12)
H1 :  μ < 12
where μ is the population mean wait time

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c)

This is a left tailed test
Using the given formulae, we get
Degrees of Freedom
df = n - 1

df = 11

t-statistic
$\newline t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}} \newline\newline\newline t=\frac{11.0833-12}{\frac{1.0836}{\sqrt{12}}}$
t-statistic = -2.930

The test statistic t = -2.930

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d)
P-value
For t = -2.930, df = 11 we find the left tailed p-value using t tables or Excel function t.dist
p-value = t.dist(-2.930, 11, TRUE)
p-value = 0.0068
The p-value = 0.0068

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e)

Decision
0.0068 < 0.01
that is

The p-value is less than α

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f)

Based on this we should Reject the null hypothesis

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g)

Conclusion   (Option 1)
The data suggest the population mean is significantly less than 12 at α = 0.01, so there is statistitically significant evidence to conclude that the population mean wait time for women with small children is less than 12

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