HW_Hyp_Tests_Part2_Solutions

Show how you verified this yes we have a random

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Unformatted text preview: othesis H0: Alternate Hypothesis H1: Significance Level α = Population Proportion = 0.56 Population Proportion "not equal" 0.56 5% The name of the TI-83/84 calculator function you will be using: 1-PropZTest Are the conditions met in order to use this statistical method? Show how you verified this: Yes. We have a random sample from a huge population, and our sample size is large enough to pass this test: np(1-p) = (300)(0.56)(1 - 0.56) = 73.92 > 10; So our Sampling Distribution for the sample proportion should be approximately normal. From your calculator’s output give the following: Test Statistic z = 1.40 Interpret what this value is telling you: Our Test Statistic is 1.40 Standard Errors above the hypothesized population proportion of 56%. P-Value = 16.28% Interpret what this value is telling you: There is a 16.28% probability of getting a Test Statistic this far away (or farther away) from the hypothesized population proportion of 56% IF the Null Hypothesis is true. Which one is your conclusion (Click in the box next to your choice): Failed to Reject H0 Reject H0 and Accept H1 Show the reason for your conclusion (to prove you weren’t just guessing): P-value > Significance Level 16.28% > 5%, so the probability is not small enough to be considered unusual. Which one is your conclusion “in context” (Click in the box next to your choice): “There is not sufficient evidence at the 0.05 significance level to support the claim that this population proportion has changed from its 1999 level. “There is sufficient evidence at the 0.05 significance level to support the claim that this population proportion has changed from its 1999 level. Page 8 of 9 b) If you are wrong about this conclusion, what type of error would it be? Type II Error. This would mean that the population proportion of all cases of tuberculosis that are from foreign-born residents is different from 56%, but we didn't catch it. This is a Missed Opportunity. c) Find a 95% confidence interval for the proportion of all cases of tuberculosis that are from foreign-born residents. The name of the TI-83/84 calculator function you will be using: 1-PropZInt State your answer in words: We are 95% confident that the population proportion of ALL cases of tuberculosis that are from foreign-born residents is between 54.456% and 65.544%. Notice that the hypothesized population proportion of 56% is in this interval, which is why it was not rejected in the Hypothesis Test. d) How large of a sample size would you need in order to have a 95% confidence level and a margin of error E = .025? Use your prior estimate for p: ˆp=x/n . Be sure to show how your answer was calculated: n = (p-hat) (1 - p-hat) (z / E)^2 = (0.6) (1 - 0.6) (1.9600 / 0.025)^2 = 1475.1744 We Round UP to: n = 1476 ================================= The End =========================== Page 9 of 9...
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This note was uploaded on 11/21/2013 for the course MTH 243 taught by Professor Dr.kim during the Winter '08 term at Southern Oregon.

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