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...


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



Blood glucose levels for obese patients have a mean of 100 with a standard deviation of 15. A<br>researcher thinks that a diet high in raw corn starch will have a negative or positive effect on<br>blood glucose levels. A sample of 30 patients who have tried the raw corn starch diet have a<br>mean glucose level of 140. Test the hypothesis that the raw cornstarch had an effect at a<br>significant level of 0.05<br>Step 1: Determine the null hypothesis<br>Ho: µ =<br>H:u +<br>Step 2: Two-tailed test<br>Step 3: Select significance level a =<br>Step 4: Do we do a Z test of T test?<br>Step 5: Find the critical value Zeri from the t Table.<br>Lvel of Confidence<br>Draw and label the graph.<br>From the t Table. Zerit fora = 0.05, ± 1.96<br>Fail to Reject<br>Reject<br>Reject<br>Note that a= 0.05 (5%) means 2.5% in each tail. This alpha<br>level indicates the Level of Confidence to be 95%.<br>a/2<br>a/2<br>Zcrit =<br>Zcrit =<br>Step 6: Calculate the Test Statistic (Zobt)<br>N =<br>X =<br>Step 7-9. State the conclusion<br>

Extracted text: 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 Ho: µ = H:u + 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 Zeri from the t Table. Lvel of Confidence Draw and label the graph. From the t Table. Zerit fora = 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%. a/2 a/2 Zcrit = Zcrit = Step 6: Calculate the Test Statistic (Zobt) N = X = Step 7-9. State the conclusion

Jun 01, 2022
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