Question

an article in a journal reports that 34% of american fathers take no responsibility for child care. a researcher claims that the figure is higher for fathers in the town of littleton. a random sample of 225 fathers from littleton, yielded 97 who did not help with child care. find the p - value for a test of the researchers claim.

a. 0.0529
b. 0.0015
c. 0.0019
d. 0.0038

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 97$ and $n=225$. So, $\hat{p}=\frac{97}{225}\approx0.4311$.

Step2: Calculate the test - statistic

The hypothesized proportion $p_0 = 0.34$. The formula for the test - statistic $z$ in a one - sample proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$.
Substitute the values:

$$
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$$

Step3: Calculate the p - value

Since this is a right - tailed test (researcher claims the proportion is higher), the p - value is $P(Z>z)$. Using the standard normal table or a calculator, $P(Z > 2.88)=1 - P(Z\leq2.88)$.
From the standard normal table, $P(Z\leq2.88)=0.9980$. So, $p - value=1 - 0.9980=0.0020$ (approximate value, more accurately using a calculator: $P(Z>2.88)\approx0.0019$)

Answer:

C. 0.0019