HW 1 STAT 252
Problem #1
A)
B) C)
D) Young Professional would be a good advertising outlet for online brokers.
Problem #2
A) Ho: u = $520 Ha: u > $520
B)
C) Assumptions: 1) Simple Random Sample (SRS): We can assume that the sample was selected by random. 2) 10% Assumption: We can assume that the sample size (n=27) is less than 10% of the population. 3) Nearly Normal: n = 27, which is less than n=30. We need to look at the boxplot and histogram to determine if the sample is normal. The diagram shows that the sample data is skewed to the left, with an outlier at 200. The sample is not normal because the diagram is not unimodal, symmetrical, and it contains an outlier.
D)
E) Since the p-value is less than the significance level at 0.05, we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the mean weekly pay for all working women is significantly greater than that for women with a high school degree.
F)
G) Results from part e) and f) agree. From the confidence interval we can conclude that we are 95% confident that the true mean weekly pay for all working women is between (603.4189 and 692.9515). AARP Bulletin’s reported average weekly pay for a woman with a high school degree is $520, which does not lie within the confidence interval.
H)
I) Yes, the conclusion would still be the same. Even though the outlier was removed, the conclusion is the same and from the confidence interval, we can conclude that we are 95% that the true mean weekly pay for all working women is between (636.9351, 693.911). The reported mean of 520 from AARP Bulletin still does not fall within the confidence interval.
Problem #3
A)
B)
C) Yes, the proportion of business students at Bayview who were involved in some type of cheating is less than that of business students (p=0.56). We are 95% confidence that the
A)
B) C)
D) Young Professional would be a good advertising outlet for online brokers.
Problem #2
A) Ho: u = $520 Ha: u > $520
B)
C) Assumptions: 1) Simple Random Sample (SRS): We can assume that the sample was selected by random. 2) 10% Assumption: We can assume that the sample size (n=27) is less than 10% of the population. 3) Nearly Normal: n = 27, which is less than n=30. We need to look at the boxplot and histogram to determine if the sample is normal. The diagram shows that the sample data is skewed to the left, with an outlier at 200. The sample is not normal because the diagram is not unimodal, symmetrical, and it contains an outlier.
D)
E) Since the p-value is less than the significance level at 0.05, we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the mean weekly pay for all working women is significantly greater than that for women with a high school degree.
F)
G) Results from part e) and f) agree. From the confidence interval we can conclude that we are 95% confident that the true mean weekly pay for all working women is between (603.4189 and 692.9515). AARP Bulletin’s reported average weekly pay for a woman with a high school degree is $520, which does not lie within the confidence interval.
H)
I) Yes, the conclusion would still be the same. Even though the outlier was removed, the conclusion is the same and from the confidence interval, we can conclude that we are 95% that the true mean weekly pay for all working women is between (636.9351, 693.911). The reported mean of 520 from AARP Bulletin still does not fall within the confidence interval.
Problem #3
A)
B)
C) Yes, the proportion of business students at Bayview who were involved in some type of cheating is less than that of business students (p=0.56). We are 95% confidence that the