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2. During the pandemic people have become more health conscious, especially in eating red meat. In March, the average consumption of red meat per person was 51 kilos. A sample of 100 persons showed that they consumed, on average, 48 kilos of red meat in April with a standard deviation of 12 kilos. Using a = 0.05, is there enough evidence to conclude that the mean consumption of red meat decreased in April.

2. During the pandemic people have become more health conscious, especially in eating red meat. In March, the average consumption of red meat per person was 51 kilos. A sample of 100 persons showed that they consumed, on average, 48 kilos of red meat in April with a standard deviation of 12 kilos. Using a = 0.05, is there enough evidence to conclude that the mean consumption of red meat decreased in April.

MATLAB: An Introduction with Applications
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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2. During the pandemic people have become more health conscious, especially in eating red meat. In March, the average consumption of red meat per person was 51 kilos. A sample of 100 persons showed that they consumed, on average, 48 kilos of red meat in April with a standard deviation of 12 kilos. Using a = 0.05, is there enough evidence to conclude that the mean consumption of red meat decreased in April.
Transcribed Image Text:2. During the pandemic people have become more health conscious, especially in eating red meat. In March, the average consumption of red meat per person was 51 kilos. A sample of 100 persons showed that they consumed, on average, 48 kilos of red meat in April with a standard deviation of 12 kilos. Using a = 0.05, is there enough evidence to conclude that the mean consumption of red meat decreased in April.
PROBLEM 1. A manufacturer claims that the average lifetime of his lightbulbs is three 3 years or 36 months. The standard deviation is 8 months. Fifty bulbs are selected, and the average lifetime is found to be 32 months, Should the manufacturer's statement be rejected at a = 0.01? SOLUTION: STEP 1: State the hypotheses Ho: H = 36 months Hạ: p # 36 months STEP 2: Level of significance a = 0.01 STEP 3: Determine the critical values and rejection region. The significance level is 0.01. the # sign in the alternative hypothesis indicates that the test is two-tailed with two rejection regions, one in each tail of the normal distribution curve of p. Because the total area of both rejections is 0.01 (Significance level), the area of rejection in each tail is 0.005, since the sample population is more than 30, z-test is to be utilized. The area to the right of u is 0.5, the area between 0 and the critical value z is 0.495. looking at the z-table the critical value is z = ±2.575. Fall to Reject the Nall Hypothesis Critical Critical Value Value (-) (+) STEP 4: State the decision rule. Reject the null hypothesis if ze > 2.575 or z. < -2.575. STEP 5: Compute the test statistic. 32- 36 z = = -3.54 8/V50 STEP 6: Make a decision. = -3.54 is less than the critical value z = -2.575 and it falls in The test statistic ze the rejection region in the left tail. Therefore, reject Ho and conclude that the average lifetime of the lightbulb is not equal to 36 months.
Transcribed Image Text:PROBLEM 1. A manufacturer claims that the average lifetime of his lightbulbs is three 3 years or 36 months. The standard deviation is 8 months. Fifty bulbs are selected, and the average lifetime is found to be 32 months, Should the manufacturer's statement be rejected at a = 0.01? SOLUTION: STEP 1: State the hypotheses Ho: H = 36 months Hạ: p # 36 months STEP 2: Level of significance a = 0.01 STEP 3: Determine the critical values and rejection region. The significance level is 0.01. the # sign in the alternative hypothesis indicates that the test is two-tailed with two rejection regions, one in each tail of the normal distribution curve of p. Because the total area of both rejections is 0.01 (Significance level), the area of rejection in each tail is 0.005, since the sample population is more than 30, z-test is to be utilized. The area to the right of u is 0.5, the area between 0 and the critical value z is 0.495. looking at the z-table the critical value is z = ±2.575. Fall to Reject the Nall Hypothesis Critical Critical Value Value (-) (+) STEP 4: State the decision rule. Reject the null hypothesis if ze > 2.575 or z. < -2.575. STEP 5: Compute the test statistic. 32- 36 z = = -3.54 8/V50 STEP 6: Make a decision. = -3.54 is less than the critical value z = -2.575 and it falls in The test statistic ze the rejection region in the left tail. Therefore, reject Ho and conclude that the average lifetime of the lightbulb is not equal to 36 months.
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