Respuesta :
Answer:
a) The null hypothesis is represented as
H₀: p = 0.28
The alternative hypothesis is given as
Hₐ: p ≠ 0.28
b) Critical value for z = 1.960
Rejection regions include
z < -1.960 and z > 1.960
c) The standardized test statistic = 1.50
d) For this question, the standardized test statistic obtained is within the non-fiction region, hence, we fail to reject the null hypothesis.
Furthermore, the p-value obtained is greater than the significance level at which the test was performed, hence, we fail to reject the null hypothesis.
e) We can conclude that there is significance evidence that the society's claim is true and the proportion of households that own a cat is not different from 0.28.
Explanation:
(a) Identify the claim and state H₀ and Hₐ.
What is the claim?
(b) Find the critical value(s) and identify the rejection region(s).
(c) Find the standardized test statistic.
(d) Decide whether to reject or fail to reject the null hypothesis.
(e) interpret the decision in the context of the original claim.
Solution
a) For hypothesis testing, the first thing to define is the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.
For this question, we want to investigate if the proportion of households that own a cat is different from the society's claim that 28% of households own a cat.
So, the null hypothesis would be that there is no significant evidence to conclude that the proportion of households that own a cat is different from the society's claim that 28% of households own a cat.
And the alternative hypothesis is that there is significant evidence to conclude that the proportion of households that own a cat is different from the society's claim that 28% of households own a cat.
Mathematically,
The null hypothesis is represented as
H₀: p = 0.28
The alternative hypothesis is given as
Hₐ: p ≠ 0.28
b) The test is a two tailed test performed at a significance level of 0.05. Using the z-distribution, the critical value for 0.05 significance level is 1.960.
Therefore, the rejection regions include
z < -1.960 and z > 1.960
c) So, we compute the z-test statistic
z = (x - μ)/σₓ
x = sample proportion of the 200 households sampled that own a cat = (66/200) = 0.33
μ = p₀ = the stamdard we are comparing against = 0.28
σₓ = standard error = √[p(1-p)/n]
where n = Sample size = 200
p = 0.33
σₓ = √[0.33×0.67/200] = 0.0332490601 = 0.03325
z = (0.33 - 0.28) ÷ 0.03325 = 1.50
d) We then obtain the p-value.
p-value (for z = 1.50, at 0.05 significance level, with a two tailed condition) = 0.133614
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 0.05
p-value = 0.133614
0.133614 > 0.05
Hence,
p-value > significance level
This means that we fail to reject the null hypothesis.
e) Failing to reject the null hypothesis means that there is no significant evidence to conclude that the proportion of households that own a cat is different from the society's claim that 28% of households own a cat.
Hence, we can conclude that there is significance evidence that the society's claim is true and the proportion of households that own a cat is not different from 0.28.
Hope this Helps!!!
