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

Question

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.

City_#1   City_#2
100   117
86   62
121   100
111   85
101   80
104   107
213   110
114   111
290   111
100   133
318   101
145   209

Use a 0.05 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.

The test statistic is

The P-value

State the conclusion for the test.

a reject, There is sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater.

b reject, There is not sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater.

c fail to reject, There is sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater.

d fail to reject, There is not sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater.

Construct a confidence interval suitable for testing the claim that the mean amount of strontium-90 from city #1 residents is greater than the mean amount from city #2 residents.

___mBq<1-2<___mBq

Does the confidence interval support the conclusion of the test?

(yes/no) because the confidence interval contains (zero, only positive values/only negative values)

Explanation / Answer

Hypothesis:

Ho: x1 - x2 = 0

Ha : x1 - x2 > 0

Test statistic:

x1 = 150.25 , x2 = 110.5 , s1 = 75.87 , s2 = 34.64 , n1 = n2 = 12

SE = sqrt[(s12/n1) + (s22/n2)]

SE = sqrt [(75.872/12) + (34.642/12 ]

= 24.17

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

= [ (150.25 - 110.5) - 0 ] / 24.17

= 1.65

Now,w e need to find p value byusing t = 1.65 , ddf = 12 + 12 -2 = 22

p value = .1131

Now, p value is greater than 0.05

So, we fail to reject the null hypothesis

fail to reject, There is not sufficient evidence to support the claim that the mean amount of strontium-90 from city #1 residents is greater.

.

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