Below is JMP output from a statistical analysis of cholesterol levels and the patient's gender.Provide the proper interpretation of the confidence interval for the difference of meancholesterol levels between genders.4t TestMALE-FEMALEAssuming unequal variancesDifference-22.283 t Ratio-1.94233Std Err Dif11.472 DF60.99917It|-52.786 Prob > tUpper CL Dif8.220 Prob >0.0567-40-2040Lower CL Dif0.9716Confidence0.99 Prob

Given : 99% confidence interval is (-52.786, 8.220)

Answered: Below is JMP output from a statistical…

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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It was reported that last year the average price of a gallon gasoline in a city X was $3.15. This year a sample of 50 gas station had an average price of $3.10 for a gallon. We assume that the population standard deviation of prices is $0.15. We are interested in determining whether this year mean price is less than last year. Perform a hypothesis test at the level of significance α=0.05.
A study was conducted to investigate drinking problems among college students. In 1983, a group of students were asked whether they had ever driven an automobile while drinking. In 1987, after the legal drinking age was raised, a different group of college students was asked the same question. The results are as shown in the table below. Test to compare the proportions of students with a drinking problem at 0.05 level of significance, 1987 versus 1983; and provide the chi-squared statistic value below (in 2 decimal places). Year Drove while drinking Yes No 1983 1,258 1,355 1987 994 1,643
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The manager of a political party wants to find out the association between the campaign spending (Large, medium and small) and the vote obtained (Majority, Average and minority). Use Chi-squared hypothesis test for association to test that there is an association between the variable in campaign spending and the variable in vote obtained with five percent level of significance. The manager uses MS excel to calculate the Chi-squared stat value and the result is 8.874. Critical value: At 1% = 13.277 :At 5% = 9.488 :At 10% = 7.779 1. State the decision rule 2. Conclusion and decision 3. State the alternative hypothesis and the null hypothesis
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In a hypothesis test for the population variance, the alternate hypothesis is the population variance doesn't equal 17.0. The significance level to be used is 0.05 and the sample size to be taken is 25. The table value(s) to use from the chi-square distribution is/are 12.401 and 39.364. 13.181 39.364 40.671
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