Listed in the data table are amounts of strontium-90 (in millbecquerels, or mBq.

Question

Listed in the data table are amounts of strontium-90 (in millbecquerels, 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 Cick the icon to view the data table of strontium-90 amounts. a. Use a 0 05 significance level to test the dam that the mean amount of strontium 90 rom city #1 residents is greater than the mean amount from dy #2 residents What are the null and alternative hypotheses? Assume that population 1 consists of amounts from city #1 kvels and population 2 consists of amounts rorn city #2 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 A O B O c O D Re ot the null hypothesis There is not suficient evidence to support the claim that the mean amount of strontium-9 om city #1 residents is greater Re ect the null hypothesis There is su cient evidence to support the claim that the mean amount of strontium-90 f om city #1 residents is greater Fail to re ect the null hypothesis There is sufficient evidence to support the claim that the mean amount of strontium 90 from city #1 residents is greater Fail to re ect the nu. hypothesis There is not sufficient evidence to support the claim that the mean amount of strontium 90 from chy #1 residents is greater b, Construct a confidence interval suitable for testing the cdaim that the mean amount of stronlum-90 from city 81 residents is greater than the mean amount from cily #2 residents mBq

Explanation / Answer

Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1< 2
Alternative hypothesis: 1 > 2

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(69.38752/12) + (37.30332/12)] = 22.7416

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (69.38752/12 + 37.30332/12)2 / { [ (69.38752 / 12)2 / (11) ] + [ (37.30332 / 12)2 / (11) ] }
DF = 267475.268519 / (14634.2268909 + 1222.45768343) = 16.868 or 17

t = [ (x1 - x2) - d ] / SE = [ (145.0833 - 110.9167) - 0 ] / 22.7416 = 1.5

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a one-tailed test, the P-value is the probability that a t statistic having 17 degrees of freedom is more extreme than 1.5.

We use the t Distribution Calculator to find P(t < 1.5)

The P-Value is 0.075979.
The result is not significant at p < 0.05.

Interpret results. Since the P-value (0.075979) is greater than the significance level (0.05), we cannot reject the null hypothesis.

(D) Fail to reject the null hypothesis. There is not sufficient evidence to support the claim.

Calculation for Confidence interval

Pooled Variance
s2p = (SS1 + SS2) / (df1 + df2) = 6206.16 / 22 = 282.1

Standard Error
s(M1 - M2) = ((s2p/n1) + (s2p/n2)) = ((282.1/12) + (282.1/12)) = 6.86

Confidence Interval
1 - 2 = (M1 - M2) ± ts(M1 - M2) = 34.1666 ± (2.07 * 6.86) = 34.1666 ± 14.220237

1 - 2 = (M1 - M2) = 34.1666, 95% CI [19.946363, 48.386837].

You can be 95% confident that the difference between your two population means (1 - 2) lies between 19.946363 and 48.386837.

145.08333 - 110.91667 = 34.16 (Yes, because this lies in the Confidence interval.)

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