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

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

(% w h o o w n e d c a t s) (n) \to (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:

H o : p = 0.50

H a : 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)

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

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