Test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.

The null and alternative hypothesis would be:

a. H0:p≥0.9H0:p≥0.9
H1:p<0.9H1:p<0.9
b. H0:p=0.9H0:p=0.9
H1:p≠0.9H1:p≠0.9
c. H0:p≤0.9H0:p≤0.9
H1:p>0.9H1:p>0.9
d. H0:μ=0.9H0:μ=0.9
H1:μ≠0.9H1:μ≠0.9
e. H0:μ≤0.9H0:μ≤0.9
H1:μ>0.9H1:μ>0.9
f. H0:μ≥0.9H0:μ≥0.9
H1:μ<0.9H1:μ<0.9

The test is:
a. left-tailed
b. two-tailed
c. right-tailed

Based on a sample of 100 people, 94% owned cats

The p-value is: (to 2 decimals)

Based on this we:

Fail to reject the null hypothesis

Reject the null hypothesis

Box 1: Select the best answer

Box 2: Select the best answer

Box 3: Enter your answer as an integer or decimal number. Examples: 3, -4, 5.5172
Enter DNE for Does Not Exist, oo for Infinity

Box 4: Select the best answer

We are given that we have to test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.

So, Null Hypothesis, [tex]H_0[/tex] : p = 0.90

Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.9

Here, the test is two tailed because we have given that to test the claim that the proportion of people who own cats is significantly different than 90% which means it can be less than 0.90 or more than 0.90.

Now, test statistics is given by;

[tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1) , where, n = sample size = 100

[tex]\hat p[/tex] = 0.94 (given)

So, Test statistics = [tex]\frac{0.94-0.90}{\sqrt{\frac{0.94(1-0.94)}{100} } }[/tex] = 1.68

Now, P-value = P(Z > 1.68) = 1 - P(Z <= 1.68)

= 1 - 0.95352 = 0.0465 ≈ 0.05 or 5%

Now, our decision rule is that;

If p-value < significance level ⇒ Reject null hypothesis

If p-value > significance level ⇒ Accept null hypothesis

Since, here p-value is less than significance level as 0.05 < 0.1, so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that proportion of people who own cats is significantly different than 90%.