Question

A real estate agent claims that more than 70% of retired couples living in apartments prefer apartment living over living in a detached house. To test this hypothesis, a random sample of 80 couples in a particular large apartment village are interviewed and 62 of them stated that they prefer apartment living.

(a) Test the agent’s claim. Make sure you state the null and alternative hypothesis, calculate the pvalue (use the p-value method), make your decision and draw your conclusion (note that no level of significance is given in this question). Also: how much evidence is there against the null hypothesis? (

b) Calculate a 95% confidence interval for p, and make sure you give a good interpretation of the resulting interval (what does the interval mean in the context of the question).

Answer 1

a) H₀: p = 0.7

H₁: p > 0.7

= 62/80 = 0.775

The test statistic z = ( - p)/sqrt(p(1 - p)/n)

= (0.775 - 0.7)/sqrt(0.7 * 0.3/80)

= 1.46

P-value = P(Z > 1.46)

= 1 - P(Z < 1.46)

= 1 - 0.9279

= 0.0721

At 5% significance level, the p-value is greater than the significance level (0.0721 > 0.05), so we should not reject the null hypothesis.

There is a very little evidence against the null hypothesis.

There is not sufficient evidence to support the agent's claim that more than 70% of retired couples living in apartments prefer apartment living over living in a detached house.

b) At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval is

+/- z_0.025 * sqrt((1 - )/n)

= 0.775 +/- 1.96 * sqrt(0.775 * (1 - 0.775)/80)

= 0.775 +/- 0.0915

= 0.6835, 0.8665

We are 95% confident that the true population proportion of retired couples living in aparatments prefer apartment living over living in a detached lies within the confidence boundaries 0.6835 and 0.8665.

Since the confidence interval contains 0.7, so we should not reject the null hypothesis.