Question

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

Answer 1

Answer:

1. H0 : p = 0.9

   H1 : p ≠ 0.9

2. The test is two tailed.

3. Reject the null hypothesis

Step-by-step explanation:

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, $H_0$ : p = 0.90

Alternate Hypothesis, $H_1$ : p $\neq$ 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;

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

                                                       $\hat p$ = 0.94 (given)

So, Test statistics = $\frac{0.94-0.90}{\sqrt{\frac{0.94(1-0.94)}{100} } }$ = 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%.