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 selecte 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. - X X More Info O A. Ho H SH2 OB. Ho H1 = H2 H: 1 H2 OC. Ho H1 = P2 OD. Ho: H1 * H2 City # 1 104 86 121 City #2 117 86 100 The test statistic is. (Round to two decimal places as needed.) 111 85 101 104 213 149 290 100 315 145 88 107 110 The P-value is (Round to three decimal places as needed.) State the conclusion for the test. 111 O A. 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. 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. 115 133 101 209 OC. 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. O D. 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. Print Done
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 selecte 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. - X X More Info O A. Ho H SH2 OB. Ho H1 = H2 H: 1 H2 OC. Ho H1 = P2 OD. Ho: H1 * H2 City # 1 104 86 121 City #2 117 86 100 The test statistic is. (Round to two decimal places as needed.) 111 85 101 104 213 149 290 100 315 145 88 107 110 The P-value is (Round to three decimal places as needed.) State the conclusion for the test. 111 O A. 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. 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. 115 133 101 209 OC. 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. O D. 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. 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.
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 A. Ho: H1SH2
H1: H1> H2
O B. Ho: H1 = H2
H1: 41 H2
City #1
O D. Ho: H1 H2
H1: 41> H2
City #2
O C. Ho: H1=#2
H1: 41> H2
104
117
86
86
121
100
The test statistic is. (Round to two decimal places as needed.)
111
85
101
88
The P-value is (Round to three decimal places as needed.)
104
107
213
110
State the conclusion for the test.
149
111
290
115
100
133
O A. 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.
315
101
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.
145
209
OC. 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.
O D. 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.
Print
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