Question

In 2012, a survey of 102 homes in a region found that 58 had overestimated market values. Suppose you want to estimate the true proportion of homes in this region with market values that are overestimated.

1. Here the population of interest is:
a. proportion of home with overestimated market value
b. all homes in the region
c. 102 homes whose market values were surveyed
d. 58 homes with overestimated market value

2. Here the sample is:
a. all homes in the region
b. proportion of home with overestimated market value
c. 58 homes with overestimated market value
d. 102 homes whose market values were surveyed

3. What is the target parameter in this problem?
a. 102 sampled homes
b. proportion of homes with overestimated market values in the region
c. proportion of homes with overestimated market values in the sample
d. no parameter is involved in this problem

4. Here, the standard error of sample proportion takes the value: [Answer to 4 decimal places]

5. Here, the margin of error for the 98% confidence interval of sample proportion is (here z∗=2.326z∗=2.326) [Answer to 4 decimal places]

6. The 98% confidence interval of proportion of homes in the region with overestimated market values:
a. (0.4656, 0.6717)
b. (0.4546, 0.6827)
c. (0.4513, 0.6859)
d. (0.4679, 0.6693)

7. Is the sample size sufficiently large to compute the above 98% confidence interval?
a. Yes, because np̂ ≥10np^≥10 and n(1−p̂ )≥10n(1−p^)≥10.
b. It is not possible to answer this question without knowing what is the number of failure here.
c. No, the sample size is actually not sufficiently large.
d. Yes, because n≥30n≥30.

Answer 1

Answer)

1)

Population of interest is all homes in the region

2)

Sample is 102 homes whose market values were surveyed

3)

Target parameter is proportion of homes with overestimated market values in the region.

4)

Point estimate (p) = 58/102 = 0.5686274509803

Standard error = √p*(1-p)/√n

N = 102

Standard error = 0.0490388274509 = 0.049

5)

First we need to check the conditions of normality

If n*p and n*(1-p) both are greater than 10 or not

N*p = 58

N*(1-p) = 44

As both are greater than 10

Conditions are met, so we can use standard normal z table to estimate the interval.

From z table, critical value for 98% confidence level is 2.326

Margin of error = z*standard error = 0.113974 = 0.114

6)

Confidence interval is given by

(P-moe, p+moe)

(0.4546, 0.6827)

7)

Yes, because np̂ ≥10np^≥10 and n(1−p̂ )≥10n(1−p^)≥10.