please help solve attached H2B. Ho: H1 + H2H1: H1 > H2O C. Ho: H1 = H2H1: H1 # H2OD. Ho: H1 = H2H1: H1 > H2City #1101City #2117The test statistic is|. (Round to two decimal places as...


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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<br>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<br>the mean amount from city #2 residents.<br>Click the icon to view the data table of strontium-90 amounts.<br>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.<br>More Info<br>A. Ho: H1 SH2<br>H1: 41> H2<br>B. Ho: H1 + H2<br>H1: H1 > H2<br>O C. Ho: H1 = H2<br>H1: H1 # H2<br>OD. Ho: H1 = H2<br>H1: H1 > H2<br>City #1<br>101<br>City #2<br>117<br>The test statistic is|. (Round to two decimal places as needed.)<br>86<br>87<br>121<br>100<br>111<br>85<br>The P-value is<br>(Round to three decimal places as needed.)<br>101<br>88<br>104<br>107<br>State the conclusion for the test.<br>213<br>110<br>150<br>111<br>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.<br>290<br>144<br>100<br>133<br>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.<br>292<br>101<br>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.<br>145<br>209<br>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.<br>Print<br>Done<br>Click to select your answer(s).<br>Save for Later<br>

Extracted 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. Print Done Click to select your answer(s). Save for Later
Jun 07, 2022
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