2244 Final- Sample Set 1 Question 46. A local wireless provider is interested in

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2244 Final- Sample Set 1 Question 46. A local wireless provider is interested in current usage patterns of potential customers to develop appealing promotional deals. They surveyed a simple random sample (SRS) of 16 po number of da conditions were satisfied, and then computed the 90% confidence interval for the mean daytime minutes to be (22.5 minutes/day, 30.5 minutes/day). tential customers to estimate the mean ytime minutes used on their cell phones. Using these data, they first confirmed necessary The wireless provider currently includes 30 daytime minutes (per day) in some of their service plans. If mean daytime usage of potential customers is less than 30 minutes/day, the company will consider lowering the number of minutes allowed per day. Using as%significance level,the company uses their survey data to test whether the population mean daytime minutes is less than 30 minutes/day. Which of the following options is the P-value for this test? Assume the necessary assumptions are reasonable. us-0.at A. B. C. D. E. P

Explanation / Answer

46.
we use Z test single for this problem because they given population standard deviation and mean
Given that,
population mean(u)=30
standard deviation, =30.5
sample mean, x =22.5
number (n)=16
null, Ho: =30
alternate, H1: <30
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 22.5-30/(30.5/sqrt(16)
zo = -0.98361
| zo | = 0.98361
critical value
the value of |z | at los 5% is 1.645
we got |zo| =0.98361 & | z | = 1.645
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : left tail - ha : ( p < -0.98361 ) = 0.16265
hence value of p0.05 < 0.16265, here we do not reject Ho
ANSWERS
---------------
null, Ho: =30
alternate, H1: <30
test statistic: -0.98361
critical value: -1.645
decision: do not reject Ho
p-value: 0.16265

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