Complete the following exercises from "Review Questions" located at the end of each chapter and put them into a Word document to submit to the instructor.

1.Chapter 9, number 9.13

9.13 Given a sample size of 36, how large does the population standard deviation have to be in order for the standard error to be (a) 1 ? (b) 2 ? (c) 5 ? (d) 100 ?

2.Chapter 10, number 10.10

10.10 Let's assume that, over the years, a paper and pencil test of anxiety yields a mean score of 35 for all incoming college freshmen. We wish to determine whether the scores of a random sample of 20 new freshmen, with a mean of 30 and a standard deviation of 10, can be viewed as coming from this population. Test at the .05 level of significance.

When using a significance level of .05, the p-value will be between 0.02 and 0.05, which is less than the level of significance of .05. In this case you would reject the null hypothesis and it can be concluded that the sample of new data is not from the original population.

3.Chapter 11, numbers 11.13 and 11.19

11.13 Reread the problem described in Question 10.5 on page 191. (a) What form should H0 and H1 take if the investigator is concerned only about salary discrimination against female members?

10.5 NULL HYPOTHESIS (H0)Once the problem has been described, it must be translated into a statistical hypoth-esis regarding some population characteristic. Abbreviated as H0, the null hypothesis becomes the focal point for the entire test procedure (even though we usually hope to reject it). In the test with SAT scores, the null hypothesis asserts that, with respect to the national average of 500, nothing special is happening to the mean score for the local population of freshmen. An equivalent statement, in symbols, reads:0:500Hwhere H0 represents the null hypothesis and is the population mean for the local freshman class.Generally speaking, the null hypothesis (H0) is a statistical hypothesis that usu-ally asserts that nothing special is happening with respect to some characteristic of the underlying population. Because the hypothesis testing procedure requires that the hypothesized sampling distribution of the mean be centered about a single number (500), the null hypothesis equals a single number (H0: 500). Furthermore, the null hypothesis always makes a precise statement about a characteristic of the population, never about a sample. Remember, the purpose of a hypothesis test is to determine whether a particular outcome, such as an observed sample mean, could have reason-ably originated from a population with the hypothesized characteristic.

(b) If this hypothesis test supports the conclusion of salary discrimination against female members, a costly class-action suit will be initiated against American colleges and universities. Under these circumstances, do you recommend using the .05 or the .01 level of significance? Why

10.5 (a) The observed difference between $80,100 and $82,500 cannot be interpreted at face value, as it could have happened just by chance. A hypothesis test permits us to evaluate the effect of chance by measuring the observed difference relative to the standard error of the mean.(b) All female members of the APA with a Ph.D. degree and a full-time teaching appointment.(c) H0: = 82,500(d) H1: 82,500(e) Reject H0 at the .05 level of significance ifz 1.96 or z -1.96(f) 80,00082,5002,4004.006000600100z(g) Reject H0 at the .05 level of significance becausez = -4.00 is more negative than -1.96.(h) The average salary of all female APA members (with a Ph.D. and a full-time teach-ing appointment) is less than $82,500.

11.19 How should a projected hypothesis test be modified if you're particularly concerned about

(a) the type I error?

(b) the type II error?

4.Chapter 12, number 12.8

12.8 In Review Question 11.12 on page 218, instead of testing a hypothesis, you might prefer to construct a confidence interval for the mean weight of all 2-pound boxes ofcandy during a recent production shift. (a) Given a population standard deviation of .30 ounce and a sample mean weight of 33.09 ounces for a random sample of 36 candy boxes, construct a 95 percent con-fidence interval (b) Interpret this interval, given the manufacturer's desire to produce boxes of candy that, on the average, exceed 32 ounces.

11.12 A production line at a candy plant is designed to yield 2-pound boxes of assorted candies whose weights in fact follow a normal distribution with a mean of 33 ounces and a standard deviation of .30 ounce. A random sample of 36 boxes from the pro-duction of the most recent shift reveals a mean weight of 33.09 ounces. (Incidentally, if you think about it, this is an exception to the usual situation where the investigator hopes to reject the null hypothesis.) (a) Describe the population being tested. (b) Using the customary procedure, test the null hypothesis at the .05 level of significance. (c) Someone uses a one-tailed test, upper tail critical, because the sample mean of 33.09 exceeds the hypothesized population mean of 33. Any comment?