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

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Test the claim that the proportion of people who own cats is significantly different than&amp;nbsp;50%&amp;nbsp;at the 0.01 significance level.&amp;nbsp;Based on a sample of 300 people,&amp;nbsp;41%&amp;nbsp;owned cats&amp;nbsp;Hint: Multiply to obtain the number of successes(%&amp;nbsp;who&amp;nbsp;owned&amp;nbsp;cats)(n)→(41%)(300)&amp;nbsp;The test statistic is: ? (to 2 decimals)&amp;nbsp;The p-value is: ? (to 4 decimals)

rogreenhoxa8 · Accepted Answer

The number of success will be:&amp;nbsp;0.41=&amp;nbsp;x300&amp;nbsp;x=0.41&amp;nbsp;⋅&amp;nbsp;300&amp;nbsp;x=123&amp;nbsp;It is necessary to test the following alternative and null hypotheses:Ho&amp;nbsp;:&amp;nbsp;p=0.50&amp;nbsp;Ha&amp;nbsp;:&amp;nbsp;p≠0.50&amp;nbsp;The z-statistic is computed as follows:&amp;nbsp;z=&amp;nbsp;p¯&amp;nbsp;−&amp;nbsp;p0p0(1&amp;nbsp;−&amp;nbsp;p0)/n=&amp;nbsp;0.41&amp;nbsp;−&amp;nbsp;0.500.50(1&amp;nbsp;−&amp;nbsp;0.50)/300=&amp;nbsp;−3.118&amp;nbsp;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.&amp;nbsp;As the p-value 0.0018 is less than 0.01 level of signfiicance, the null hypothesis is rejected.&amp;nbsp;The test statistic is: -3.12(to 2 decimals)&amp;nbsp;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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