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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?

User JaggenSWE
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1 Answer

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Answer:

1 : Construction of hypothesis:

2 :Test Statistic given below

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.

User Snehal Patel
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