Exercise 14

(a) State the decision rule, (b) compute the pooled estimate of the population variance, (c) compute the test statistic, (d) state your decision about the null hypothesis, and (e) estimate the p-value.

The null and alternate hypotheses are:

A random sampleof 15 observations from the first population revealed a sample meanof 350 and a sample standard deviationof 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .10 significance level, is there a difference in the population means?

Exercise 18

Use the five-step hypothesis testingprocedure for the following exercises.

The Tampa Bay (Florida) Area Chamber of Commerce wanted to know whether the mean weekly salary of nurses was larger than that of school teachers. To investigate, they collected the following information on the amounts earned last week by a sample of school teachers and nurses.

School teachers ($) 845 826 827 875 784 809 802 820 829 830 842 832

Nurses ($) 841 890 821 771 850 859 825 829

Is it reasonable to conclude that the mean weekly salary of nurses is higher? Use the .01 significance level. What is the p-value?

Exercise 22

The federal government recently granted funds for a special program designed to reduce crime in high-crime areas. A study of the results of the program in eight high-crime areas of Miami, Florida, yielded the following results.

Number of Crimes by Area

A B C D E F G H

Before 14 7 4 5 17 12 8 9

After 2 7 3 6 8 13 3 5

Has there been a decrease in the number of crimes since the inauguration of the program? Use the .01 significance level. Estimate the p-value.

Exercise 26

A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. A random sample of 50 regular-coffee drinkers showed a mean of 4.35 cups per day, with a standard deviation of 1.20 cups per day. A sample of 40 decaffeinated-coffee drinkers showed a mean of 5.84 cups per day, with a standard deviation of 1.36 cups per day. Use the .01 significance level. Compute the p-value.

See attached file for full problem description.

Please see the attachment.

Chapter 11

Exercise 14

(a) State the decision rule, (b) compute the pooled estimate of the population variance, (c) compute the test statistic, (d) state your decision about the null hypothesis, and (e) estimate the p-value.

The null and alternate hypotheses are:

A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .10 significance level, is there a difference in the population means?

Ans

The null and alternate hypotheses are:

H0: ( the mean of first and second sample)

H1 :

Test statistic is

T = where

The decision rule is if calculated value of T is greater than critical value reject the null hypothesis

Sample N Mean StDev SE Mean

1 15 350.0 12.0 3.1

2 17 342.0 15.0 3.6

Difference = mu (1) – mu (2)

Estimate for difference: 8.00000

90% CI for difference: (-0.22634, 16.22634)

T-Test of difference = 0 (vs not =):

T-Value = 1.65

P-Value = …

The solution gives complete details of comparison of population means based on different types of student t tests. The answer also include null hypothesis, alternative hypothesis, test statistics, decision rule, critical value and p value.