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

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

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.