A publisher reports that 49I% 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 200200 found that 42B% of the readers owned a personal computer. Determine the P-value of the test statistic. Round your answer to four decimal places.

We are given that a publisher reports that 49% of their readers own a personal computer. A random sample of 200 found that 42% of the readers owned a personal computer.

And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;

Null Hypothesis, [tex]H_0[/tex] : p = 0.49 {means that the percentage of readers who own a personal computer is same as reported 63%}

Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.49 {means that the percentage of readers who own a personal computer is different from the reported 63%}

The test statistics we will use here is;

T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ~ N(0,1)

where, p = actual % of readers who own a personal computer = 0.49

[tex]\hat p[/tex] = percentage of readers who own a personal computer in a

sample of 200 = 0.42

n = sample size = 200

So, Test statistics = [tex]\frac{0.42 -0.49}{\sqrt{\frac{0.42(1- 0.42)}{200} } }[/tex]

= -2.00

Now, P-value of test statistics is given by = P(Z > -2.00) = P(Z < 2.00)

= 0.9773 .