Question

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37) A researcher would like to determine whether a new tax on cigarettes has had any effect on people's behavior. During the year before the tax was imposed, stores located in rest areas on the state thruway reported selling an average of u = 410 packs per day with s = 60. The distribution of daily sales was approximately normal. For a sample of n = 9 days following the new tax, the researcher found an average of M = 386 packs per day for the same stores. Is the sample mean sufficient to conclude that there was a significant change in cigarette purchases after the new tax. Use an a = .05. Conducta hypothesis test. Step 1: State your hypotheses in complete sentences and relating it to the research question (2 points each/4 points totall Ho: Step 2: Set your criteria. Include a drawing of your critical regions in the box below. (5 points) one-tail or 2 tail-test df Critical value(s)

Answer 1

37.

Given that,
population mean(u)=410
sample mean, x =386
standard deviation, s =60
number (n)=9
null, Ho: μ=410
alternate, H1: μ!=410
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.306
since our test is two-tailed
reject Ho, if to < -2.306 OR if to > 2.306
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =386-410/(60/sqrt(9))
to =-1.2
| to | =1.2
critical value
the value of |t α| with n-1 = 8 d.f is 2.306
we got |to| =1.2 & | t α | =2.306
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.2 ) = 0.2645
hence value of p0.05 < 0.2645,here we do not reject Ho
ANSWERS
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null, Ho: μ=410
alternate, H1: μ!=410
test statistic: -1.2
critical value: -2.306 , 2.306
decision: do not reject Ho
p-value: 0.2645
we do not have enough evidence to support the claim that mean is 410 packs per day for the same stores.
type 2 error is possible because fails to reject the null hypothesis.