Based on a sample of 300 people,

Hint: To get the number of successes, multiply

The test statistic is: ? (to 2 decimals)

The p-value is: ? (to 4 decimals)

sanuluy
2020-11-05
Answered

Test the claim that the proportion of people who own cats is significantly different than $50\mathrm{\%}$ at the 0.01 significance level.

Based on a sample of 300 people,$41\mathrm{\%}$ owned cats

Hint: To get the number of successes, multiply

$(\mathrm{\%}\text{}who\text{}owned\text{}cats)(n)\to (41\mathrm{\%})(300)$

The test statistic is: ? (to 2 decimals)

The p-value is: ? (to 4 decimals)

Based on a sample of 300 people,

Hint: To get the number of successes, multiply

The test statistic is: ? (to 2 decimals)

The p-value is: ? (to 4 decimals)

You can still ask an expert for help

rogreenhoxa8

Answered 2020-11-06
Author has **109** answers

The number of success will be:

$0.41=\text{}\frac{x}{300}$

$x=0.41\text{}\cdot \text{}300$

$x=123$

The following null and alternative hypotheses need to be tested:

$Ho\text{}:\text{}p=0.50$

$Ha\text{}:\text{}p\ne 0.50$

The z-statistic is computed as follows:

$z=\text{}\frac{\overline{p}\text{}-\text{}{p}_{0}}{\sqrt{{p}_{0}(1\text{}-\text{}{p}_{0})/n}}=\text{}\frac{0.41\text{}-\text{}0.50}{\sqrt{0.50(1\text{}-\text{}0.50)/300}}=\text{}-3.118$

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)

The following null and alternative hypotheses need to be tested:

The z-statistic is computed as follows:

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)

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