A grinding machine will be qualified for a particular task if it can be shown to pro-duce less than 8% defective parts. In a random sample of 300 parts, 12 were defective.4.You may use Minitab or other software for this problem, but you must show enough stepsthat it is clear that you understand the procedure(s) you are using.(a)0.08? Perform an appropriate hypothesis test.Do the data provide evidence that the probability of a defective part is belowCalculate a 99% confidence interval for the probability of a defective part. Use(b)the quick-and-dirty confidence interval method to test whether the probability of a de-fective part is 0.08.(c)Do your answers in parts (a) and (b) agree about whether the probability isbelow 0.08?

Answered: A grinding machine will be qualified…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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After the political ad campaign, pollsters check the mayor's positives. They test the hypothesis that the ads produced no change against the alternative that the positives are now above 47% and find a P-value of 0.283. Which conclusion is appropriate? Explain. Choose the correct answer below. O A. There is a 28.3% chance that natural sampling variation could produce poll results at least as far above 47% as these if there is really no change in public opinion. There is a 28.3% chance that the poll they conducted is correct. O B. O C. There is a 71.7% chance that the ads worked. O D. There is a 28.3% chance that the ads worked.
23.Statistical power provides the same information as a p-value. True or False? 24. A statistically significant effect (i.e., p < .05) will always be practically meaningful. True or False ? 26. Rejecting the null hypothesis means that the sample outcome is very unlikely to have occurred if H0 is true. True or False ?
A student at a university wants to determine if the proportion of students that use iPhone's is less than 0.38. The hypotheses for this scenario are as follows. Null Hypothesis: p = 0.38, Alternative Hypothesis: p < 0.38. If the student randomly samples 29 other students and finds that 12 of them use iPhone's, what is the test statistic and p-value?
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You hear on the local news that for the city of Kalamazoo, the proportion of people who support President Trump is 0.42. However, you think it is less than 0.42. The hypotheses you want to test are Null Hypothesis: p ≥ 0.42, Alternative Hypothesis: p < 0.42. You take a random sample around town and calculate a p-value for your hypothesis test of 0.7963. What is the appropriate conclusion? Conclude at the 5% level of significance. Question 4 options: 1) We did not find enough evidence to say the proportion of people who support President Trump is larger than 0.42. 2) We did not find enough evidence to say a significant difference exists between the proportion of people who support President Trump and 0.42 3) The proportion of people who support President Trump is significantly less than 0.42. 4) We did not find enough evidence to say the proportion of people who support…
hopes that the results of a trial period will enable her to conclude with a level of significance of 0.05 that the new procedure increases the average profit of installing a carpet. What will be the consequences if she makes a Type I error on her hypothesis test? A Type I error would mean that Carpetland implements a new procedure that actually decreases profits. A Type I error would mean that Carpetland implements a new procedure that actually increases profits. A Type 1 error would mean that Carpetland does not implement a new procedure that increases profits. A Type I error would mean that Carpetland does not implement a new procedure that decreases profits.. A Type I error would mean that Carpetland implements a new procedure that actually does not increase profits. A Type I error would mean that Carpetland implements a new procedure that actually does not decrease profits.
Evaluate the significance of the p-value. You can think of the p-value as the probability that the null hypothesis is true. A value greater than 0.05 indicates that in more than 1 in 20 (or more than 5%) repetitions of an experiment of the same size, the observed deviation from predicted valuescould have been obtained by chance, even if the null hypothesis is actually true; the data are therefore not significant for rejecting the null hypothesis. Statisticians have arbitrarily selected the 0.05 p-value as the boundarybetween rejecting or not rejecting the null hypothesis. A p-value of less than 0.05 means that you can consider the deviation to be significant, and you can reject the null hypothesis.
3
A researcher claims that 45% of U.S. adults think that the IRS is not aggressive enough in pursuing people who cheat on their taxes. You believe the true percentage is more than 45%. To test this, you take a simple random sample of 400 U.S. adults and find that 208 of them say the IRS is not aggressive enough. Test at .05 significance. Round to the fourth Ho: P HA: P How many successes were there? Test Statistic: = ४ .45 .45 P-value: Did something significant happen? Significance Happened Select the Decision Rule: Reject the Null There is Oenough evidence to conclude that more than 45% of U.S. adults think that the IRS is not aggressive enough in pursuing people who cheat on their taxes. ✓
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A researcher claims that 45% of U.S. adults think that the IRS is not aggressiv who cheat on their taxes. You believe the true percentage is more than 45%. random sample of 400 U.S. adults and find that 196 of them say the IRS is no .10 significance. Round to the fourth as needed What's the minimum population size required? How many successes were there? Ho: P HA: P Test Statistic: P-value: 0 =
Takeout Food A magazine article reported that 14% of adults buy takeout food every day. A fast-food restaurant owner surveyed 180 customers and found that 31 said that they purchased takeout food every day. At a=0.01, is there evidence to believe the article's claim? Use the critical value method with tables. Part 1 of 5 (a) State the hypotheses and identify the claim. This hy is a two-tailed V test. Part (b th e critical vaiue, separate them with co Critic Part: 2 / 5 Part 3 of 5 (c) Compute the test value. Always round z score values to at least two decimal places. z=

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