Question

According to a recent study, the mean number of people in a family (a married couple) in Canada is 4.80. A sociologist thinks this may not be the correct value for the city that she lives in. She examines a large apartment complex and interviews 100 randomly selected families (married couples) in the complex and obtains a mean of 4.64 persons per family. It is also known that the population standard deviation, σ, can be assumed to be 0.796. You will test the sociologist’s claim using a two-sided test under three different methods and three different levels of significance. First, state the null and alternative hypothesis in symbols and words. Then test the hypothesis, make your decision and draw your conclusion using the following methods:

(a) A 98% confidence interval

(b) Classical testing with α = 5%

(c) The p-value method with a 10% level of significance. Also: how much evidence is there against the null hypothesis (very little, mild, strong, very strong or extremely strong)?

Answer 1

H₀: = 4.8

H₁: 4.8

a) At 98% confidence interval the critical value is z_0.01 = 2.33

The 98% confidence interval is

+/- z_0.01 *

= 4.64 +/- 2.33 * 0.796/

= 4.64 +/- 0.185

= 4.455, 4.825

Since the confidence interval contains 4.8, so we should not reject the null hypothesis.

There is very little evidence to reject the null hypothesis.

b) The test statistic z = ()/()

= (4.64 - 4.8)/(0.796/)

= -2.01

At alpha = 0.05, the critical values are z_0.025 = +/- 1.96

Since the test statistic value is less than the negative critical value (-2.01 < -1.96), so we should reject the null hypothesis.

There is mild evidence to reject the null hypothesis.

c) P-value = 2 * P(Z < -2.01)

= 2 * 0.0222 = 0.0444

Since the p-value is less than the significance level (0.0444 < 0.1), so we should reject the null hypothesis.

There is very strong evidence to reject the null hypothesis.