Question

A publisher reports that 54% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually over the reported percentage. A random sample of 330 found that 60% of the readers owned a personal computer. Is there sufficient evidence at the 0.05 level to support the executive's claim?

step 1 of 5 : State the null and alternative hypotheses.

Step 2 of 5 :  Find the value of the test statistic. Round your answer to two decimal places.

Step 3 of 5 : Specify if it is one tailed or two tailed

Step 4 of 5 :  Find the P-value of the test statistic. Round your answer to four decimal places.

Step 5 of 5 : Make the decision to reject or fail to reject the null hypothesis.

Answer 1

Here, we have to use one sample z test for the population proportion.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: the percentage is not actually over the reported percentage.

Alternative hypothesis: Ha: the percentage is actually over the reported percentage.

H0: p = 0.54 versus Ha: p < 0.54

This is an upper tailed test.

We are given

Level of significance = α = 0.05

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

n = sample size = 330

p̂ = x/n = 0.60

p = 0.54

q = 1 - p = 0.46

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.60 – 0.54)/sqrt(0.54*0.46/330)

Z = 2.1869

Test statistic = 2.1869

Critical value = 1.6449

(by using z-table)

P-value = 0.0144

(by using z-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the percentage is actually over the reported percentage.

There is sufficient evidence at the 0.05 level to support the executive's claim.