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. Use a

0.050.05

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.

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Click the icon to view the data table of strontium-90 amounts.
City_#1 City_#2

107 117

86 71

121 100

118 85

101 87

104 107

213 110

129 111

290 146

100 133

325 101

145 209

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.

A.

Upper H 0H0:

mu 1μ1equals=mu 2μ2

Upper H 1H1:

mu 1μ1greater than>mu 2μ2

B.

Upper H 0H0:

mu 1μ1not equals≠mu 2μ2

Upper H 1H1:

mu 1μ1greater than>mu 2μ2

C.

Upper H 0H0:

mu 1μ1less than or equals≤mu 2μ2

Upper H 1H1:

mu 1μ1greater than>mu 2μ2

D.

Upper H 0H0:

mu 1μ1equals=mu 2μ2

Upper H 1H1:

mu 1μ1not equals≠mu 2μ2

The test statistic is

nothing.

(Round to two decimal places as needed.)The P-value is

nothing.

(Round to three decimal places as needed.)

State the conclusion for the test.

A.

RejectReject

the null hypothesis. There

isis

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

B.

Fail to rejectFail to reject

the null hypothesis. There

isis

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

C.

RejectReject

the null hypothesis. There

is notis not

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

D.

Fail to rejectFail to reject

the null hypothesis. There

is notis not

Click to select your answer(s).

Answer #1

For Sample 1 :

∑x = 1839

∑x² = 351187

n1 = 12

Mean , x̅1 = Ʃx/n = 1839/12 = 153.2500

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(351187-(1839)²/12)/(12-1)] = 79.4070

For Sample 2 :

∑x = 1377

∑x² = 172281

n2 = 12

Mean , x̅2 = Ʃx/n = 1377/12 = 114.7500

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(172281-(1377)²/12)/(12-1)] = 36.0180

--

Null and Alternative hypothesis: **C.**

Ho : µ1 ≤ µ2

H1 : µ1 > µ2

Test statistic:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (153.25 -
114.75)/√(79.407²/12 + 36.018²/12) = **1.53**

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 15.3425 = 15

p-value = T.DIST.RT(1.5296, 15) = **0.073**

Decision:

p-value > α, **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.

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