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30 April, 13:00

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?

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  1. 30 April, 13:04
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    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.
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