Listed in the data table are amounts of strontium-90 (in millibecquerels, or mBg. per gram of calcium) in a simple random sample of baby teeth obtained from residents in two cities. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use a 0.01 significance level to test the claim that the mean amount of strontium-90 from city #1 residents is greater than the mean amount from city #2 residents. E Click the icon to view the data table of strontium-90 amounts. What are the null and alternative hypotheses? Assume that population 1 consists of amounts from city #1 levels and population 2 consists of amounts from city #2. More Info O B. Ho: H1 P2 H: H > H2 OD. Ho: H1 =P2 O A. Ho: H1 SP2 H: H >H2 OC. Ho: H1 =P2 H: H, H2 H: H >H2 City #2 City #1 101 86 121 117 The test statistic is . (Round to two decimal places as needed.) 87 100 111 85 88 The P-value is (Round to three decimal places as needed.) 101 104 107 110 State the conclusion for the test. 213 150 111 O A. Reject the null hypothesis. There is sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater. 290 100 292 144 O B. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater. 133 101 c. Reject the null hypothesis. There is not sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater. O D. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater. 145 209 Print Done
Listed in the data table are amounts of strontium-90 (in millibecquerels, or mBg. per gram of calcium) in a simple random sample of baby teeth obtained from residents in two cities. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use a 0.01 significance level to test the claim that the mean amount of strontium-90 from city #1 residents is greater than the mean amount from city #2 residents. E Click the icon to view the data table of strontium-90 amounts. What are the null and alternative hypotheses? Assume that population 1 consists of amounts from city #1 levels and population 2 consists of amounts from city #2. More Info O B. Ho: H1 P2 H: H > H2 OD. Ho: H1 =P2 O A. Ho: H1 SP2 H: H >H2 OC. Ho: H1 =P2 H: H, H2 H: H >H2 City #2 City #1 101 86 121 117 The test statistic is . (Round to two decimal places as needed.) 87 100 111 85 88 The P-value is (Round to three decimal places as needed.) 101 104 107 110 State the conclusion for the test. 213 150 111 O A. Reject the null hypothesis. There is sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater. 290 100 292 144 O B. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater. 133 101 c. Reject the null hypothesis. There is not sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater. O D. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater. 145 209 Print Done
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Listed in the data table are amounts of strontium-90 (in millibecquerels, or mBq, per gram of calcium) in a simple random sample of baby teeth obtained from residents in two cities. Assume that the two samples are independent simple random
samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use a 0.01 significance level to test the claim that the mean amount of strontium-90 from city #1 residents is greater than
the mean amount from city #2 residents.
Click the icon to view the data table of strontium-90 amounts.
What are the null and alternative hypotheses? Assume that population 1 consists of amounts from city #1 levels and population 2 consists of amounts from city #2.
More Info
A. Ho: H1 SH2
H1: 41> H2
B. Ho: H1 + H2
H1: H1 > H2
O C. Ho: H1 = H2
H1: H1 # H2
OD. Ho: H1 = H2
H1: H1 > H2
City #1
101
City #2
117
The test statistic is|. (Round to two decimal places as needed.)
86
87
121
100
111
85
The P-value is
(Round to three decimal places as needed.)
101
88
104
107
State the conclusion for the test.
213
110
150
111
A. Reject the null hypothesis. There is sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater.
290
144
100
133
B. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater.
292
101
C. Reject the null hypothesis. There is not sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater.
145
209
D. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater.
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