Week 4 homework assignment THIS is the assignment TEMPLATE that you please use Psychology 520

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1 Homework Week 4 Your Name (without credentials) Grand Canyon University Psychology 520: Graduate Statistics Christopher Smith Assignment Due Date

2 1. Chapter 9 Question 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 1 ? 2 ? 5 ? 100 ? Answer( s ) and commentary ( please show work ): ( a ) SE = sample ¿¿ √ ¿ Population Standard Deviation ¿ SE =1 1= σ √ 36 σ = 6 ( b ) SE = sample ¿¿ √ ¿ Population Standard Deviation ¿ SE =2 2= σ √ 36 σ = 12 ( c ) SE = sample ¿¿ √ ¿ Population Standard Deviation ¿ SE =5 5= σ √ 36 σ = 30

3 ( d ) SE = sample ¿¿ √ ¿ Population Standard Deviation ¿ SE =100 100= σ √ 36 σ = 6 00 2. Chapter 10 Question 10.10

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4 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. Answer( s ) and commentary ( please show work ): This is a two tailed test T - test = ´ X − μ s √ n = 30 − 35 10 √ 20 = -2.24 P – value = 0.0375 Reject the null hypothesis. There is sufficient evidence to claim that the mean is not equal to 35 at 0.05 significance level.

5 3. Chapter 11 Question 11.13 11.13 Reread the problem described in Question 10.5 on page 191. What form should H0 and H1 take if the investigator is concerned only about salary discrimination against female members? 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? Progress Check *10.5 According to the American Psychological Association, members with a doctorate and a full-time teaching appointment earn, on the average, $82,500 per year, with a standard deviation of $6,000. An investigator wishes to determine whether $82,500 is also the mean salary for all female members with a doctorate and a full-time teaching appointment. Salaries are obtained for a random sample of 100 women from this population, and the mean salary equals $80,100. Someone claims that the observed difference between $80,100 and $82,500 is large enough by itself to support the conclusion that female members earn less than male members. Explain why it is important to conduct a hypothesis test. The investigator wishes to conduct a hypothesis test for what population? What is the null hypothesis, H0? What is the alternative hypothesis, H1? Specify the decision rule, using the .05 level of significance. Calculate the value of z. (Remember to convert the standard deviation to a standard error.) What is your decision about H0? Using words, interpret this decision in terms of the original problem. Answers on page 433. 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: μ = 82,500 (d) H1: μ ≠ 82,500 (e) Reject H0 at the .05 level of significance if z ≥ 1.96 or z ≤ –1.96 (f) z = - 2400 / 600 = - 4 (g) Reject H0 at the .05 level of significance because z = –4.00 is more negative than –

6 1.96. (h) The average salary of all female APA members (with a Ph.D. and a full-time teaching appointment) is less than $82,500. Answer( s ) and commentary ( please show work ): ( a ) The mean pay for female members holding a PhD and a full-time teaching position should equal $82,500, which is the null hypothesis H0. The alternative hypothesis (H1) is that female members who hold a PhD and are employed full-time as teachers make an average income of less than $82,500. This is a result of the investigator's exclusive focus being on wage discrimination against female participants. ( b ) It is advised to utilize the.01 level of significance if the results of this hypothesis test support the conclusion that female members are subjected to salary discrimination. This is due to the fact that a class-action lawsuit is an expensive and serious matter, and stronger proof opposing the null hypothesis is provided by a greater degree of significance. The probability of making a Type I error, or rejecting the null hypothesis when it is true, is merely 1%, according to a.01 threshold of significance.

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7 4. Chapter 11 Question 11.19 11.19 How should a projected hypothesis test be modified if you're particularly concerned about the type I error? the type II error? Answer( s ) and commentary ( please show work ): ( a ) Type 1 error Reduce the Sample size if too large, when it results a rejection of null hypothesis. This happens a high significance level, so in order to reduce the significance level 0.01 to 0.001 ( b ) Type 2 error Increase the sample size if too small, when it results a fail to reject a null hypothesis. So in order to correct increase the sample size.

8 5. Chapter 12 Question 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 of candy during a recent production shift. 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 confidence interval. 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 production of

9 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.) Describe the population being tested. Using the customary procedure, test the null hypothesis at the .05 level of significance. 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? Answer( s ) and commentary ( please show work ): ( a ) SE = sample ¿¿ √ ¿ Population Standard Deviation ¿ SE = 0.30 √ 36 = 0.05 The critical value for a 95% confidence level in a standard normal distribution is 1.96 from the Z – distribution table. ConfidenceInterval = 33.09 ± ( 1.96 )( 0.05 ) ConfidenceInterval = 33.09 ± 0.098 Therefore, the confidence interval will be: (32.992, 33.188) ( b ) We are 95% certain that the true population mean weight of candy boxes ranges between 32.992 and 33.188 ounces.

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10 6. Answer( s ) and commentary ( please show work ): Please leave blank. Thank you !! ?

11 7. Answer( s ) and commentary ( please show work ): Please leave blank. Thank you !! ?

12 8. Answer( s ) and commentary ( please show work ): Please leave blank. Thank you !! ?

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13 9. Answer( s ) and commentary ( please show work ): Please leave blank. Thank you !! ?

14 10. Answer( s ) and commentary ( please show work ): Please leave blank. Thank you !! ?

15 11. Please leave this question blank. Please leave blank. Thank you !! ?

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16 References Type your references here using hanging indent and double line spacing (under “Paragraph” on the Home toolbar ribbon). See your APA Manual ( seventh edition ) and the resources in the APA section of Resources for reference formatting.

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