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

sanuluy 2020-11-05 Answered
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)(41%)(300)
The test statistic is: ? (to 2 decimals)
The p-value is: ? (to 4 decimals)
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Expert Answer

rogreenhoxa8
Answered 2020-11-06 Author has 109 answers
The number of success will be:
0.41= x300
x=0.41  300
x=123
The following null and alternative hypotheses need to be tested:
Ho : p=0.50
Ha : p0.50
The z-statistic is computed as follows:
z= p¯  p0p0(1  p0)/n= 0.41  0.500.50(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)
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