Question

A publisher reports that 48%48% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually under the reported percentage. A random sample of 110110 found that 40%40% of the readers owned a personal computer. Is there sufficient evidence at the 0.050.05 level to support the executive's claim?

Step 1 of 7:

State the null and alternative hypotheses.

Step 2 of 7:

Find the value of the test statistic. Round your answer to two decimal places.

Step 3 of 7:

Specify if the test is one-tailed or two-tailed

Step 4 of 7:

Determine the P-value of the test statistic. Round your answer to four decimal places.

Step 5 of 7:

Identify the value of the level of significance.

Answer 1

Given that,
possibile chances (x)=44
sample size(n)=110
success rate ( p )= x/n = 0.4
success probability,( po )=0.48
failure probability,( qo) = 0.52
null, Ho:p=0.48
alternate, H1: p<0.48
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.4-0.48/(sqrt(0.2496)/110)
zo =-1.679
| zo | =1.679
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =1.679 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: left tail - Ha : ( p < -1.67944 ) = 0.04653
hence value of p0.05 > 0.04653,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.48
alternate, H1: p<0.48
test statistic: -1.679
critical value: -1.645
decision: reject Ho
p-value: 0.04653
we have enough evidence to support the claim that the percentage of 48% is actually under the reported percentage.