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
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Author:Amos Gilat
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Chapter1: Starting With Matlab
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Question
II
Sh) with Zcrit of
Therefore,
(a = .01).
%66
Level of Confidence
Fail to Reject
Reject
Reject
0.5%
0.5%
Zcrit =
-2.576
bubiy
Z crit = +2.576
3.33
%3D
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
%3D
= 11 : °H
+ri : "H
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 Zerii from the t Table.
Level of Confidence
Draw and label the graph.
Fail to Reject
From the t Table. Zerit for a= 0.05, + 1.96
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)
= X
%3D
= N
= rl
Step 7-9. State the conclusion
Transcribed Image Text:II Sh) with Zcrit of Therefore, (a = .01). %66 Level of Confidence Fail to Reject Reject Reject 0.5% 0.5% Zcrit = -2.576 bubiy Z crit = +2.576 3.33 %3D 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 %3D = 11 : °H +ri : "H 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 Zerii from the t Table. Level of Confidence Draw and label the graph. Fail to Reject From the t Table. Zerit for a= 0.05, + 1.96 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) = X %3D = N = rl Step 7-9. State the conclusion
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