Assignment 4

# The sample data are compatible with the cracking rate

This preview shows pages 2–4. Sign up to view the full content.

evidence to show that the cracking rate has decreased from 25%. The sample data are compatible with the cracking rate of 25%. Z = = -4.0166 P-value = Pr (Z < -4.016) = 0.0000 Assuming a significance level of 0.05, 0.0000 < alpha, there is evidence to say that the cracking rate is no longer 25%. The evidence shows that the changes made to the casting process may have helped reduce the cracking rate.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
c) The two hypothesis tests had different outcomes. There was not enough evidence to reject the null hypothesis in the first test, but we rejected it in the second. The lesson to be learned is that the p-value decreases as the sample size increases, meaning that our tests will be more statistically significant. d) Yes, the results would be the same, because the z-stat will just be positive instead of the negative value that we obtained from the first test with the same p-value. Question 4 – Jeopardy! winnings a) The sampling distribution of the mean winnings (in \$) is bimodal with two peaks, somewhat symmetrical, and has an outlier (at the point \$35,201). Mean = \$18,459.25 St. Dev = \$6,543.70 b) 95% CI: (15396.74, 21521.76) c) No, the 95% CI for µ is an interval that has a 95% chance of containing the mean winnings of the entire population, NOT the winnings of 95% of similar winners. d) I. H o : μ = μ o H a : μ μ o μ o = 34064 This is a two-sided alternative because the question asks whether or not the population mean differs from Jennings’ average (alternative hypothesis has a “not equal” sign), which means that μ o can either be larger or smaller than μ . II. t = = -10.6647 P-value = 2 x Pr (t >|t-stat|) < 0.01. P-value = 2 x < 0.01 P-value < 0.02. At the 5% significance level, 0.02 < alpha, therefore we have significant evidence to prove that the mean winnings of the population are different from Ken Jennings’ mean winnings. III. The 95% CI from part b) does support the conclusion of our hypothesis test. The CI we found was (15396.74, 21521.76). Ken Jennings’ mean winnings are not inside this interval, which means that we reject the null hypothesis. e) In this situation, Type I error would mean that Ken Jennings’ average of \$34,064 actually represented the population mean winnings, but this was rejected in the hypothesis test. Type II
error would mean that the previous statement was actually false, but we accepted the hypothesis. The only error that could have happened in part d) was the Type I error.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern