Question

A random sample of 1,496 respondents of a major metropolitan area was questioned about a number of issues. When asked to agree or disagree with the statement “Hand guns should be outlawed”, 650 respondents agreed. Researchers could thus construct a confidence interval for the proportion of the residents of this metropolitan area who support banning hand guns (use the 95% confidence level). Use this example to answer questions 19 to 23.
19. What’s the proportion of respondents in the sample who agreed “Hand guns should be outlawed”?

20. What would be the Z score(s) associated with the confidence level used in this example?

21. Based on the sample information, what’s the standard error?

22. Using the sample information to estimate the population proportion in the example, what’s the margin of error?

23. What conclusion can we make for this example?

According to data from the 2018 General Social Survey (GSS 2018), the average number of years of education of the 2345 adults in the U.S. sample is 13.73, with a standard deviation of 2.974. Compared to the national average of 13.26 years of education in 2000, researchers are wondering if the national education level had increased during these years. Do a hypothesis testing with α=0.05. Use this example to answer questions 24 to 27.
24. What’s the null hypothesis in this case?
A. The average number of years of education in the U.S. adult population did not change much from 2000 to 2018.
B. The average number of years of education in the U.S. adult population was equal to 13.73 in 2018.
C. The average number of years of education for the GSS 2018 sample is no different from 13.26, the national average in 2000.
D. The average number of years of education in the U.S. adult population had increased from 2000 to 2018.

25. What’s the alternative hypothesis in this case?
A. The average number of years of education in the U.S. adult population had changed since 2000.
B. The average number of years of education for the GSS 2018 sample is different from 13.26, the national average in 2000.
C. The average number of years of education in the U.S. adult population in 2018 was higher than that in 2000.
D. The average number of years of education in the U.S. adult population had decreased from 2000 to 2018.

26. Which of the following statements about this example is correct?
A. This is a two-tailed test and you have two rejection regions.
B. Since the sample size is large, we cannot use the normal distribution as the sampling distribution.
C. This is a one-tailed test and the rejection region is on the left side of the sampling distribution.
D. The rejection region is on the right side of the sampling distribution.

27. What conclusion can we draw for this example?
A. There is no enough evidence to reject the null hypothesis.
B. We can be 90% confident that the average number of years of education in the U.S. adult population had increased since 2000.
C. The average number of years of education in the U.S. adult population had increased since 2000.
D. There is a significant difference between 2018 and 2000 in terms of the national education level in the U.S. adult population.

Are U-Albany students more likely to approve gun control than adults in the U.S.? According to a research report, on a scale from 1 to 10, the mean approval of gun control in the U.S. is 7.8. The mean approval of gun control in a random sample of 26 U-Albany students is 8.3, with the standard deviation of 2.2. Use α = 0.01 for the hypothesis testing. Questions 28 to 32 are based on this example.
28. What would be the H0 for the example?
A. U-Albany students are equally likely to approve gun control than adults in the U.S.
B. The likelihood of approving gun control among U-Albany students is different from that of the U.S. adults.
C. There is no difference between the 26 U-Albany students and all U.S. adults in terms of their attitudes toward gun control.
D. U-Albany students are less likely to approve gun control than adults in the U.S.

29. What would be the t critical value(s)?

30. What’s the standard error based on the sample information?

31. What is the t obtained value?

32. What conclusion can we make for this example?
A. We cannot reject the null hypothesis that the mean approval of gun control among the 26 U-Albany students is 8.3.
B. There is no enough evidence to reject the null hypothesis that the mean approval of gun control among U-Albany students is 8.3.
C. We cannot reject the null hypothesis that the mean approval of gun control among all U-Albany students is 7.8.
D. We can reject H0 and accept H1 that the mean approval of gun control among all U-Albany students is larger than 7.8.

In an election exit poll (N = 1,768), 898 respondents said they voted for Candidate A. Is Candidate A going to win the election (more than 50% of votes needed to win)? Use α = 0.1 for a hypothesis testing. Questions 33 to 38 are based on this example.

33. What’s the proportion of respondents in the poll voted for Candidate A?

34. What would be the H1 for the example?
A. The percentage of respondents in the poll voted for Candidate A is different from 50%.
B. The proportion of all voters in the election who would vote for Candidate A is more than 0.5.
C. The percentage of respondents in the poll voted for Candidate A is different from 50.8%.
D. The proportion of all voters in the election who would vote for Candidate A is more than 0.51.

35. What would be the Z critical value(s)?

36. What’s the standard error based on the sample information?

37. What is the Z obtained value?

38. What conclusion can we make for this example?

(Optional) Using the International Social Survey Program data we find women in the U.S. (N = 654) spend, on average, 3.53 hours per day doing housework, with the standard deviation of 1.91, and women in Sweden (N = 639) spend, on average, 2.87 hours per day doing housework, with the standard deviation of 1.85. Researchers want to test whether American women spend more time than Swedish women on housework? Questions 39 to 42 are based on this example.
39. What is the null hypothesis?

40. What would be the critical value(s) for α = 0.05?

41. What would be the z-obtained value?

42. What conclusion can you make for this example? Briefly explain with statements and numbers.

kindly, I need them done.

Answer 1

19. What’s the proportion of respondents in the sample who agreed “Hand guns should be outlawed”?

p^=sample proportion of respondents in the sample who agreed “Hand guns should be outlawed”

=x/n=650/1496=0.434492

20. What would be the Z score(s) associated with the confidence level used in this example?

Z critical value in excel

==NORM.S.INV(0.975)

=1.959964

z critical for 95%=1.96

21. Based on the sample information, what’s the standard error?

standard error=sqrt(p^*(1-p^)/n

=sqrt(0.434492*(1-0.434492)/1496)

SE=0.01281576

22. Using the sample information to estimate the population proportion in the example, what’s the margin of error?

margin of error=Z crit*SE

=1.959964*0.01281576

= 0.02511843

23. What conclusion can we make for this example?

95% confidence interval for p is

sample proportion-margin of error,sample proportion+margin of error

0.434492-0.02511843,0.434492+0.02511843

0.4093736,0.4596104

0.4093736<p<0.4596104

we are 95% confident that the true  proportion of respondents in the sample who agreed “Hand guns should be outlawed”lies in between  0.4093736and 0.4596104