Question

The StatCrunch data set for this question contains the data measurements described in Question 11. (H₀ : µ₁ - µ₂ ≤ 0
H_A : µ₁ - µ₂ > 0) Assume that the two samples are dawn from independent, normally distributed populations that have different standard deviations. Use this data set and the results from Question 11 to calculate the p-value for the hypothesis test. Round your answer to three decimal places; add trailing zeros as needed.

The p-value = [S90PValue].

Using the result from Question 12, identify the appropriate decision for the hypothesis test in Question 11, along with its interpretation. Use α = 0.01.

Fail to reject H₀. There is insufficient evidence to support the original claim that the mean amount of Strontium-90 collected from City 1 residents' teeth is greater than the mean amount collected from the residents of City 2.

Reject H₀. There is insufficient evidence to support the original claim that the mean amount of Strontium-90 collected from City 1 residents' teeth is greater than the mean amount collected from the residents of City 2.

Fail to reject H₀. There is insufficient evidence to reject the original claim that the mean amount of Strontium-90 collected from City 1 residents' teeth is greater than the mean amount collected from the residents of City 2.

Reject H₀. There is sufficient evidence to support the original claim that the mean amount of Strontium-90 collected from City 1 residents' teeth is greater than the mean amount collected from the residents of City 2.

city1    city 2

104   117   ""
86   73   ""
121   100   ""
119   85   ""
101   84   ""
104   107   ""
213   110   ""
144   111   ""
290   105   ""
100   133   ""
275   101   ""
145   209   ""

Answer 1

Given that,
mean(x)=150.1667
standard deviation , s.d1=70.25
number(n1)=12
y(mean)=111.25
standard deviation, s.d2 =34.7017
number(n2)=12
null, Ho: u1 <= u2
alternate, H1: u1 > u2
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =2.718
since our test is right-tailed
reject Ho, if to > 2.718
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =150.1667-111.25/sqrt((4935.0625/12)+(1204.20798/12))
to =1.7206
| to | =1.7206
critical value
the value of |t α| with min (n1-1, n2-1) i.e 11 d.f is 2.718
we got |to| = 1.72055 & | t α | = 2.718
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:right tail - Ha : ( p > 1.7206 ) = 0.05665
hence value of p0.01 < 0.05665,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 <= u2
alternate, H1: u1 > u2
test statistic: 1.7206
critical value: 2.718
decision: do not reject Ho
p-value: 0.05665
we do not have enough evidence to support the claim that City 1 residents' teeth is greater than the mean amount collected from the residents of City 2.