A new post-surgical treatment is being compared with a standard treatment. Seven subjects receive the new treatment, while seven others (the controls) receive the standard treatment. The recovery times, in days, are given below. Treatment: Control: 12 13 15 19 20 21 21 18 23 24 30 32 35 39 Let µx represent the population mean for the control patients and let uy represent the population mean for the treated patients. Find a 98% confidence interval for the difference µy – Hy . Round down the degrees of freedom to the nearest integer and round the answers to three decimal places. The 98% confidence interval is (|

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.3: Measures Of Spread
Problem 1GP
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A new post-surgical treatment is being compared with a standard treatment. Seven subjects receive the new treatment, while seven
others (the controls) receive the standard treatment. The recovery times, in days, are given below.
Treatment:
12 13 15
19 20 21 21
Control:
18 23
24 30 32 35 39
Let ux represent the population mean for the control patients and let uy represent the population mean for the treated patients. Find
a 98% confidence interval for the difference ux - Hy . Round down the degrees of freedom to the nearest integer and round the
answers to three decimal places.
The 98% confidence interval is
Transcribed Image Text:A new post-surgical treatment is being compared with a standard treatment. Seven subjects receive the new treatment, while seven others (the controls) receive the standard treatment. The recovery times, in days, are given below. Treatment: 12 13 15 19 20 21 21 Control: 18 23 24 30 32 35 39 Let ux represent the population mean for the control patients and let uy represent the population mean for the treated patients. Find a 98% confidence interval for the difference ux - Hy . Round down the degrees of freedom to the nearest integer and round the answers to three decimal places. The 98% confidence interval is
In an experiment to test the effectiveness of a new sleeping aid, a sample of 12 patients took the new drug, and a sample of 14 patients
took a commonly used drug. Of the patients taking the new drug, the average time to fall asleep was 27.8 minutes with a standard
deviation of 5.2 minutes, and for the patients taking the commonly used drug the average time was 32.5 minutes with a standard
deviation of 4.1 minutes. Can you conclude that the mean time to sleep is less for the new drug? Find the P-value and state a
conclusion.
Sinc v (Click to select)
we
(Click to select) : conclude that the mean time to sleep is less for the new drug.
0.025 < P < 0.05
P< 0.0005
0.01 < P < 0.025
P> 0.4
0.001 < P < 0.005
0.05 < P < 0.1
0.1 < P < 0.25
0.25 < P < 0.4
0.0005 < P < 0.001
0.005 < P < 0.01
Transcribed Image Text:In an experiment to test the effectiveness of a new sleeping aid, a sample of 12 patients took the new drug, and a sample of 14 patients took a commonly used drug. Of the patients taking the new drug, the average time to fall asleep was 27.8 minutes with a standard deviation of 5.2 minutes, and for the patients taking the commonly used drug the average time was 32.5 minutes with a standard deviation of 4.1 minutes. Can you conclude that the mean time to sleep is less for the new drug? Find the P-value and state a conclusion. Sinc v (Click to select) we (Click to select) : conclude that the mean time to sleep is less for the new drug. 0.025 < P < 0.05 P< 0.0005 0.01 < P < 0.025 P> 0.4 0.001 < P < 0.005 0.05 < P < 0.1 0.1 < P < 0.25 0.25 < P < 0.4 0.0005 < P < 0.001 0.005 < P < 0.01
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