Question

Test the claim that the proportion of people who own cats is significantly different than \(50\%\) at the 0.01 significance level.
Based on a sample of 300 people, \(41\%\) owned cats
Hint: To get the number of successes, multiply
\((\%\ who\ owned\ cats)(n)\rightarrow(41\%)(300)\)
The test statistic is: ? (to 2 decimals)
The p-value is: ? (to 4 decimals)

Answer 1

The number of success will be:
\(0.41=\ \frac{x}{300}\)
\(x=0.41\ \cdot\ 300\)
\(x = 123\)
The following null and alternative hypotheses need to be tested:
\(Ho\ :\ p = 0.50\)
\(Ha\ :\ p \neq 0.50\)
The z-statistic is computed as follows:
\(z=\ \frac{\bar{p}\ -\ p_{0}}{\sqrt{p_{0}(1\ -\ p_{0})/n}}=\ \frac{0.41\ -\ 0.50}{\sqrt{0.50(1\ -\ 0.50)/300}}=\ -3.118\)
The required P-value can be obtained by using the standard normal table. For this case, first, find the area under the standard normal curve to the left of the z score of -3.12 (approx to two decimal places). The required area is 0 .000904. Now, this area should be doubled to reflect the nature of a two-tailed test that is 0.0018. Hence, the required P-value is 0.0018.
As the p-value 0.0018 is less than 0.01 level of signfiicance, the null hypothesis is rejected.
The test statistic is: -3.12(to 2 decimals)
The p-value is: 0.0018 (to 4 decimals)