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Choose alternative hypothesis as what is desired to be concluded or proved or tested. Then choose critical value and define decision rule according to the inequality sign in alternative hypothesis. Find test statistic and see if it satisfies decision rule.

8. At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average more than $80 a day in tips.” Assume the standard deviation of the population distribution is $3.24. Over the first 35 days she was employed at the restaurant, the mean daily amount of her tips was $84.85.

a) At the .01 significance level, can Ms. Brigden conclude that she is earning an average of more than $80 in tips?

a. State the null and alternate hypothesis.

H0: µ ≤ 80

Ha: µ > 80

b. State the decision rule.

Here we use z-test. Given significance level of 0.01, upper critical value (because Ha is saying mean is more than) is z = 2.3263

So, decision rule is Reject H0 if test statistic is greater than 2.3263

c. Compute the value of the test statistic.

Z = (xbar - µ)/(σ/sqrt(n)) = (84.85-80)/(3.24/sqrt(35)) = 8.85586

d. What is your decision regarding H0?

Reject H0.

e. What is the p-value? Interpret it.

P-value = 0.000

This means probability of obtaining observed statistic is almost 0 if average was less than or equal to 80.

12. The management of white industries is considering a new method of assembling its golf cart. The present method requires 42.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes.

a) Using the .10 level of significance, can we conclude that the assembly time using the new method is faster?

a. State the null and alternate hypothesis.

H0: µ ≥ 42.3

Ha: µ < 42.3

b. State the decision rule.

Since standard deviation is estimated from sample, we’d use t-test. For t-test, degree of freedom = n-1 = 24-1 = 23. Lower critical value (because of Ha is saying mean is less than) of t with d.f. 23 and significance level 0.1 is t = -1.31946

So, decision rule is Reject H0 if test statistic is less than -1.31946

c. Compute the value of the test statistic.

t-statistic = (40.6 - 42.3)/(2.7/sqrt(24)) = -3.0845

d. What is your decision?

Reject H0.

24. A recent article in USA Today reported that a job awaits only one in three new college graduates. The major reasons given were an overabundance of college graduates and a weak economy. A survey of 200 recent graduates from your school revealed that 80 students had jobs.

a) At the .02 significance level, can we conclude that a larger proportion of students at your school have jobs? Use the five-step hypothesis-testing procedure in answering this question.

H0: p ≤ 0.3333

Ha: p > 0.3333

Decision rule: At 0.02 significance level, upper critical value of z is 2.0537. So, decision rule is Reject H0 if z > 2.0537

Observed p = 80/200 = 0.4

So, z-statistic = (0.4-0.3333)/sqrt(0.3333*(1-0.3333)/200) = 2

Decision: Fail to reject H0.

Conclusion: There isn’t sufficient evidence to conclude that a larger proportion of students at my school have jobs.

6. Mary Jo Fitzpatrick is the vice president for Nursing Services at St. Luke’s Memorial Hospital. Recently, she noticed in the job postings for nurses that those that are unionized seem to offer higher wages. She decided to investigate and gathered the following information. (Refer to table in textbook, pg. 327).

Union 20.75 2.25 40

Non-union 19.80 1.90 45

a) Would it be reasonable for her to conclude that union nurses earn more? Use the .02 significance level.

H0: µ1 ≤ µ2

Ha: µ1 > µ2

We’d use two-sample z-test, as the sample size is sufficiently large for both samples. At 0.02 significance level, critical z = 2.0537. So decision rule is that Reject H0 if z>2.0537

z-statistic = (20.75-19.80)/sqrt(2.25^2/40+1.90^2/45) = 2.0891

Decision: Reject H0.

Conclusion: Union nurses earn more than non-union nurses.

b) What is the p-value? Use the five-step hypothesis-testing procedure to answer this question.

p-value = 0.0183

16. A recent study compared the time spent together by single and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the .01 significance level, can we conclude that the single-earner couples on average spend more time watching television together? There were 15 single-earner and 12 dual-earner couples studied.

a) Use the five-step hypothesis-testing procedure to answer this question.

H0: µ1 ≤ µ2

Ha: µ1 > µ2

We’d use t-test for independent samples assuming equal variance. Degree of freedom is 15+12-2 = 25. Upper critical value of t at 0.01 significance level and d.f. of 25 is t = 2.4851. So, decision rule is Reject H0 if t>2.4851

t-statistic = (61-48.4)/sqrt((14*15.5^2+11*18.1^2)/(14+11)*(1/15+1/12)) = 1.9488

Decision: Can’t reject H0.

Conclusion: There isn’t sufficient evidence to conclude that single-earner couples on average spend more time watching television together.

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: refer to table on page 343.

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

a) 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.

H0: µd ≤ 0 (average change (before minus after) is negative or zero, increase in crime)

Ha: µd > 0 (average change (before minus after) is positive, decrease in crime)

We’d use paired sample t-test. Degree of freedom = 8-1 = 7. At 0.01 level, upper critical value is t = 2.99795. Decision rule is to Reject H0 if t > 2.99795

Average difference (before-after) = 3.625, standard deviation of difference = 4.8385

t-statistic = (3.625-0)/(4.8385/sqrt(8)) = 2.1191

Decision: Can’t reject H0.

Conclusion: There isn’t sufficient evidence to conclude that there has been a decrease in crime.

p-value = 0.0359

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