Blood glucose levels for obese patients have a mean of 100 with a standard deviation of 15. A esearcher 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 glucose level of 140. Test the hypothesis that the raw cornstarch had an effect at a mean significant level of 0.05

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
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Reject
Reject
0.5%
0.5%
Zcrit
-2.576
3.33
Zcrit = +2.576
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
esearcher 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
:
Step 2: Two-tailed test
Step 3: Select significance level a = 0.05
Step 4: Do we do a Z test of T test? T+est
Step 5: Find the critical value Zori from the t Table.
Level of Confidence
Draw and label the graph.
From the t Table. Zerit for a= 0.05, ± 1.96
Fail to Reject
Reject
Reject
Note that a= 0.05 (5%) means 2.5% in each tail. This alpha
level indicates the Level of Confidence to be 95%.
Zcrit =
Step 6: Calculate the Test Statistic (Zobt)
N= 30
00l =
UNS
Step 7-9. State the conclusion
%3D
Transcribed Image Text:Reject Reject 0.5% 0.5% Zcrit -2.576 3.33 Zcrit = +2.576 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 esearcher 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 : Step 2: Two-tailed test Step 3: Select significance level a = 0.05 Step 4: Do we do a Z test of T test? T+est Step 5: Find the critical value Zori from the t Table. Level of Confidence Draw and label the graph. From the t Table. Zerit for a= 0.05, ± 1.96 Fail to Reject Reject Reject Note that a= 0.05 (5%) means 2.5% in each tail. This alpha level indicates the Level of Confidence to be 95%. Zcrit = Step 6: Calculate the Test Statistic (Zobt) N= 30 00l = UNS Step 7-9. State the conclusion %3D
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