▾ Part 1 of 4 A pharmaceutical company claims their new diabetes medication results in less variance in a patient's glucose level than if the patient were on no medication at all. An endocrinologist wishes to test this claim. She divides participants randomly into two groups. Group A consists of 20 diabetics who received the medication; group B consists of 26 diabetics who received a placebo. After two weeks, the blood sugar level of each patient in each group was measured with the following results (in mg/dL): Group A: 77.8, 229.4, 199.9, 110.1, 180.2, 116.1, 139.7, 171.1, 37.4, 158.1, 88.4, 195.5, 246.1, 142.4, 178.1, 105.5, 179.6, 146.1, 78.8, 123.7 Group B: 124.5, 130.1, 136, 162.8, 113.4, 72.8, 142.6, 50.3, 179.8, 197, 230, 194.3, 171, 109.3, 114.4, 107.5, 114.7, 195.3, 127.7, 126.4, 85.6, 166.7, 182.3, 113.5, 216.7, 162.8 Perform a hypothesis test using a 6% level of significance to test the pharmaceutical company's claim. Step 1: State the null and alternative hypotheses. 0² Ho: H₂: EVV (So we will be performing a left-tailed Y Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of test statistics. We will use a(n) F dfA=19 and denominator degrees of freedom dfB = 25 A = 145.2 Step 3: Find the p-value of the test statistic. SA= 2898.71 P(FSV 1.45070 ✓✓test.) distribution with numerator degrees of freedom p-value 0.1897 Question Help: D Post to forum Submit Part F= 0.1897 x 1.45070 28 143.37 S8== 1998.14 X Part 3 of 4
▾ Part 1 of 4 A pharmaceutical company claims their new diabetes medication results in less variance in a patient's glucose level than if the patient were on no medication at all. An endocrinologist wishes to test this claim. She divides participants randomly into two groups. Group A consists of 20 diabetics who received the medication; group B consists of 26 diabetics who received a placebo. After two weeks, the blood sugar level of each patient in each group was measured with the following results (in mg/dL): Group A: 77.8, 229.4, 199.9, 110.1, 180.2, 116.1, 139.7, 171.1, 37.4, 158.1, 88.4, 195.5, 246.1, 142.4, 178.1, 105.5, 179.6, 146.1, 78.8, 123.7 Group B: 124.5, 130.1, 136, 162.8, 113.4, 72.8, 142.6, 50.3, 179.8, 197, 230, 194.3, 171, 109.3, 114.4, 107.5, 114.7, 195.3, 127.7, 126.4, 85.6, 166.7, 182.3, 113.5, 216.7, 162.8 Perform a hypothesis test using a 6% level of significance to test the pharmaceutical company's claim. Step 1: State the null and alternative hypotheses. 0² Ho: H₂: EVV (So we will be performing a left-tailed Y Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of test statistics. We will use a(n) F dfA=19 and denominator degrees of freedom dfB = 25 A = 145.2 Step 3: Find the p-value of the test statistic. SA= 2898.71 P(FSV 1.45070 ✓✓test.) distribution with numerator degrees of freedom p-value 0.1897 Question Help: D Post to forum Submit Part F= 0.1897 x 1.45070 28 143.37 S8== 1998.14 X Part 3 of 4
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter4: Equations Of Linear Functions
Section4.5: Correlation And Causation
Problem 11PPS
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