Question

A publisher reports that 49% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 200 found that 42% of the readers owned a personal computer. Determine the P-value of the test statistic.

Answer 1

0.0239

Explanation:

A publisher reports that 49% of their readers own a personal computer.

Claim : . A marketing executive wants to test the claim that the percentage is actually different from the reported percentage.

`H_0:p = 0.49\\H_a:p\\eq 0.49`

A random sample of 200 found that 42% of the readers owned a personal computer.

No. of people owned a personal computer =
`42\% * 200`

=
`(42)/(100) * 200`

=
`84`

`x = 84\\n = 200`

We will use one sample proportion test

`\widehat{p}=(x)/(n)`

`\widehat{p}=(84)/(200)`

`\widehat{p}=0.42`

Formula of test statistic =
`\frac{\widehat{p}-p}{\sqrt{(p(1-p))/(n)}}`

=
`-1.98`

Now refer the p value from the z table

p value =0.0239

Hence The p value of test statistic is 0.0239