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The conclusion is that there is sufficient evidence to support the claim that the percentage of vacant housing units in their district differs from the percentage of vacant housing units in the country.
Given that the total number of a housing units in her country is 550 and the number of vacant housing units in her country is 130.
The Null hypothesis is p₀=13%=0.13, n=550 and x=130
Significance level α=0.05
Claims given: percentage other than 13%
The claim is anyone of the null hypothesis or the alternative hypothesis. The null hypothesis shows that the proportion of the population is equal to the value stated in the claim. If the null hypothesis is a claim, then the alternative hypothesis states the opposite of the null hypothesis.
H₀:p=13%=0.13
Hₐ:p/0.13
I'm interested in testing population proportion claims, so I'll use z-text with proportions.
Conditions: Random sample, 10% condition, pass / fail condition
Sample: I'm happy because the housing units were randomly selected. 10% Condition: We are happy because the sample of 550 homes in your county is less than 10% of the population.
Pass/ fail condition: Satisfied because both np₀ and n(1-p₀) are both at least 10.
np₀=550(0.13)=71.5≥10
n(1-p₀)=550(1-0.13)=478.5≥10
Therefore, we conclude that all the conditions are met.
The sample percentage is the number of successes divided by the sample size.
Determines the value of the test statistic:
The P-value is the probability of getting a test statistic value or a more extreme value if the null hypothesis is true. Use the normal probability table in the appendix to determine the P-value.
P=P(Z<−7.41 or Z>7.41)
P=2P(Z>7.41)
P=2(0)
P=0
If the P-value is less than the significance level α, the null hypothesis is rejected.
P<0.05⇒ Reject H₀
Therefore, there is ample evidence to support the claim that the proportion of vacant homes in the area is different from the proportion of vacant homes in the country.
Learn more about Standard normal distribution from here brainly.com/question/13383035
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