Of the 420 respondents in a Christmas tree market survey, 42% had no children at home and 58% had at least one child at home. The corresponding figures for the most recent census are 48% with no children and 52% with at least one child.

1. Test the null hypothesis that the telephone survey technique has a probability of selecting a household with no children that is equal to the value obtained by the census.

2. Give the z statistic (rounded to two decimal places) and the P-value. What do you conclude?

1 : Construction of hypothesis:

2 : Test Statistic given below

Step-by-step explanation:

Critical value = 48% = 48/100 = 0.48

1 : Construction of hypothesis:

H₀ : P = 0.48

H₁ : P ≠ 0.48

Sample proportion = 0.42

Variance = p * (1-p) / n

putting values = 0.48 (0.52) / 420

= 0.000594=0.0006

Taking sqrt[0.0006] = 0.025

2:

Test Statistic

Z = 0.42 - 0.48) / 0.025 = - 2.4

level of significance is 0.10

We conclude that households with no children from 48% then telephone survey technique has a probability of selecting a household with no children that is equal to the value obtained by the census.