Question

2. In a random sample of 857 families with multiple children, it was found that 533 of them prefer Charmin brand toilet paper. Test the claim that the proportion is larger than 0.61 using a 0.05 significance level.

a. Give the critical region and the value of the test statistic.

b. Give the decision and conclusion.

Answer 1

Part a

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 proportion is 0.61.

Alternative hypothesis: Ha: the proportion is larger than 0.61

H0: p = 0.61 versus Ha: p > 0.61

This is a upper tailed test.

We are given

Level of significance = α = 0.05

Critical value = 1.6449

(by using z-table)

Critical region: Reject H0 when z > 1.6449

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

x = number of items of interest = 533

n = sample size = 857

p̂ = x/n = 533/857 = 0.621936989

p = 0.61

q = 1 - p = 0.39

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.621936989 - 0.61)/sqrt(0.61*0.39/857)

Z = 0.7165

Test statistic = 0.7165

Part b

P-value = 0.2369

(by using z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that the proportion is larger than 0.61.