Listed in the data table are amounts of strontium-90 (in becquerels or mBq, per

Question

Listed in the data table are amounts of strontium-90 (in becquerels 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. Complete parts (a) and (b) below BEE Click the icon to view the data table of strontium-90 amounts. a. U a 0.10 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. se 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. O A. H H H2 o: O B. Ho: H1 H2 H1: H1 >H2 H1: H1 H2 O C. Ho: H H2 O D. Ho: 1 s H2 H1: H H2 H1: H1 H2 The test statistic is (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed State the conclusion for the test. O A. 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 B. 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 C. 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. O D. 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.

Explanation / Answer

x1 = 151.33 , x2 = 111.83 , s1 = 78.38 , s2 = 33.83 , n1 = n2 = 12

a) H0 : mu1 = mu2

Ha: mu1 > mu2

Test statistic:

SE = sqrt [(s12/n1) + (s22/n2)]
SE = sqrt [(78.38^2/12) + (33.83^2/12)]

= 24.64

t = [ (x1 - x2) - d ] / SE

= [(151.33 - 111.83) - 0] / 24.64

= 1.60

Now, we need to find p value using t = 1.60 , df = 12+12 -2 = 22

p value = .062

Answer is Option D)

b)

Confidence interval for 90% with df = 22 t value = 1.321

CI = (x1 - x2) + / - t * sqrt ( s1^2/n1 + s2^2/n2)

= (151.33 - 111.83 ) + /- 1.321 * sqrt ((78.38^2/12) + (33.83^2/12))

= (6.94 , 72.05)

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