Question

The serious delinquency rate measures the proportion of homes that are well behind on their mortgage payments. A real estate investor is trying to decide in which of two neighborhoods in Phoenix to buy an investment property. He is trying to decide if the serious delinquency rate differs between the two neighborhoods. He took a sample of 70 homes in neighborhood 1 and found 14 homes were seriously delinquent. He took a sample of 80 homes in neighborhood 2 and 10 homes were seriously delinquent. Specify the null and alternate hypotheses to determine if the delinquency rates in the two neighborhoods are equal and make a conclusion at the 10% significance level. How should the delinquency rate influence the investor’s decision?

Answer 1

We conclude that the delinquency rates in the two neighborhoods are equal.

Explanation:

We are given that a real estate investor is trying to decide in which of two neighborhoods in Phoenix to buy an investment property.

He took a sample of 70 homes in neighborhood 1 and found 14 homes were seriously delinquent. He took a sample of 80 homes in neighborhood 2 and 10 homes were seriously delinquent.

Let
`p_1` = delinquency rate in the neighborhood 1.

`p_2` = delinquency rate in the neighborhood 2.

So, Null Hypothesis,
`H_0` :
`p_1-p_2` = 0 {means that the delinquency rates in the two neighborhoods are equal}

Alternate Hypothesis,
`H_A` :
`p_1-p_2\\eq` 0 {means that the delinquency rates in the two neighborhoods are different}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. =
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ~ N(0,1)

where,
`\hat p_1` = proportion of homes that were seriously delinquent in neighborhood 1 =
`(14)/(70)` = 0.2

`\hat p_2` = proportion of homes that were seriously delinquent in neighborhood 2 =
`(10)/(80)` = 0.125

`n_1` = sample of homes in neighborhood 1 = 70

`n_2` = sample of homes in neighborhood 2 = 80

So, test statistics =
`\frac{(0.2 -0.125)-(0)}{\sqrt{(0.2(1-0.2))/(70)+(0.125(1-0.125))/(80) } }`

= 1.241

The value of z test statistics is 1.241.

Now, at 10% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the delinquency rates in the two neighborhoods are equal.