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In: Statistics and Probability

11. (10 pts) A large furniture store has begun a new ad campaign on local television....

11. (10 pts) A large furniture store has begun a new ad campaign on local television. Before the campaign, the long-term average daily sales were $24,819. A random sample of 40 days during the new ad campaign gave a sample mean daily sale average of = $25,910. Does this indicate that the population mean daily sales are now more than $24,819? Use a 1% level of significance. Assume  = $1917. (a) State the null and alternate/research hypotheses. What is the level of significance? Will you use a left-tailed, right tailed, or two-tailed test? (b) Identify the sampling distribution you will use: What is the value of the test statistic? (c) Find (or estimate) the P-value. Use MS Excel or table. (d) Based on your answers for parts (a) to (c), will you reject or fail to reject the null hypothesis? (e) Interpret your test result/conclusion in one complete sentence only.

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Expert Solution

11.

Given that,
population mean(u)=24819
standard deviation, σ =1917
sample mean, x =25910
number (n)=40
null, Ho: μ=24819
alternate, H1: μ>24819
level of significance, α = 0.01
from standard normal table,right tailed z α/2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 25910-24819/(1917/sqrt(40)
zo = 3.599
| zo | = 3.599
critical value
the value of |z α| at los 1% is 2.326
we got |zo| =3.599 & | z α | = 2.326
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : right tail - ha : ( p > 3.599 ) = 0
hence value of p0.01 > 0, here we reject Ho
ANSWERS
---------------
a.
null, Ho: μ=24819
alternate, H1: μ>24819
b.
test statistic: 3.599
c.
critical value: 2.326
p-value: 0
d.
decision: reject Ho
e.
we have enough evidence to support the claim that the population mean daily sales are now more than $24,819.

orchestra answered 2 years ago
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