es a significant difference between the population mean and t with Zcrit of sample mean of IQ test scores. The pills do work. ( α-01), %3D Therefore, 99% Level of Confidence Fail to Reject Reject Reject 0.5% 0.5% Zcrit = -2.576 3.33 Z crit = +2.576 %3D Zobt Let's Practice Two-Tailed Z Test Blood glucose levels for obese patients have a mean of 100 with a standard deviation of 15. A researcher thinks that a diet high in raw corn starch will have a negative or positive effect on blood glucose levels. A sample of 30 patients who have tried the raw corn starch diet have a mean glucose level of 140. Test the hypothesis that the raw cornstarch had an effect at a significant level of 0.05 Step 1: Determine the null hypothesis H, : µ = Step 2: Two-tailed test Step 3: Select significance level a = Step 4: Do we do a Z test of T test? Step 5: Find the critical value Zcrii from the t Table. Draw and label the graph. From the t Table. Zerit for a = 0.05, ± 1.96 Note that a= 0.05 (5%) means 2.5% in each tail. This alpha level indicates the Level of Confidence to be 95%. Level of Confidence Fail to Reject Reject Reject a/2 2 Zarit =- Z crit = Step 6: Calculate the Test Statistic (Zobt) N = Step 7-9. State the conclusion II
es a significant difference between the population mean and t with Zcrit of sample mean of IQ test scores. The pills do work. ( α-01), %3D Therefore, 99% Level of Confidence Fail to Reject Reject Reject 0.5% 0.5% Zcrit = -2.576 3.33 Z crit = +2.576 %3D Zobt Let's Practice Two-Tailed Z Test Blood glucose levels for obese patients have a mean of 100 with a standard deviation of 15. A researcher thinks that a diet high in raw corn starch will have a negative or positive effect on blood glucose levels. A sample of 30 patients who have tried the raw corn starch diet have a mean glucose level of 140. Test the hypothesis that the raw cornstarch had an effect at a significant level of 0.05 Step 1: Determine the null hypothesis H, : µ = Step 2: Two-tailed test Step 3: Select significance level a = Step 4: Do we do a Z test of T test? Step 5: Find the critical value Zcrii from the t Table. Draw and label the graph. From the t Table. Zerit for a = 0.05, ± 1.96 Note that a= 0.05 (5%) means 2.5% in each tail. This alpha level indicates the Level of Confidence to be 95%. Level of Confidence Fail to Reject Reject Reject a/2 2 Zarit =- Z crit = Step 6: Calculate the Test Statistic (Zobt) N = Step 7-9. State the conclusion II
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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