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. 0.0038
b. 0.0019
c. 0.0629
d. 0.0015

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{97}{225}\approx0.4311$. The population proportion $p = 0.34$, and $n=225$.

Step2: Calculate the test - statistic

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

$$
LATEXBLOCK0
$$

Step3: Find the p - value

Since this is a right - tailed test (the researcher claims the proportion is higher), the p - value is $P(Z>2.88)$.
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 $p - value=1 - 0.9980 = 0.0020\approx0.0019$ (due to rounding differences in intermediate steps).

Answer:

B. 0.0019