Question

Test the claim that the proportion of people who own cats is smaller than 30% at the 0.005 significance level.

The null and alternative hypothesis would be:

H0:μ≥0.3H0:μ≥0.3
H1:μ<0.3H1:μ<0.3

H0:μ≤0.3H0:μ≤0.3
H1:μ>0.3H1:μ>0.3

H0:p≤0.3H0:p≤0.3
H1:p>0.3H1:p>0.3

H0:p≥0.3H0:p≥0.3
H1:p<0.3H1:p<0.3

H0:μ=0.3H0:μ=0.3
H1:μ≠0.3H1:μ≠0.3

H0:p=0.3H0:p=0.3
H1:p≠0.3H1:p≠0.3

The test is:

right-tailed

left-tailed

two-tailed

Based on a sample of 700 people, 24% owned cats

The test statistic is:  (to 2 decimals)

The p-value is:  (to 2 decimals)

Based on this we:

• Fail to reject the null hypothesis
• Reject the null hypothesis

Answer 1

Solution:

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 of people who own cats is 30%.

Alternative hypothesis: H_a: The proportion of people who own cats is smaller than 30%.

H₀: p ≥ 0.3

H₁: p < 0.3

This is a left-tailed or lower-tailed test.

The test is: left-tailed

We are given

Level of significance = α = 0.005

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 = 700

p̂ = x/n = 0.24

p = 0.3

q = 1 - p = 0.7

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.24 - 0.3)/sqrt(0.3*0.7/700)

Z = -3.4641

Test statistic = -3.4641

The test statistic is: -3.46

P-value = 0.0003

(by using z-table)

The p-value is: 0.00

P-value < α = 0.005

So, we reject the null hypothesis

Reject the null hypothesis

There is sufficient evidence to conclude that the proportion of people who own cats is smaller than 30%.