Question

a child psychologist says that the average time that working mothers spend talking to their children is 11 minutes per day, on average. suppose a random sample of 25 working mothers spend an average of 11.5 minutes per day talking with their children. assume prior research suggests the population standard deviation is 2.3 minutes. can we say that the psychologists claim is true at α = 0.02? (use the # option for h₀ and if z is positive, change its sign to negative before looking up p - value)

Explanation:

Step1: State the hypotheses

Let $\mu$ be the population - mean time working mothers spend talking to their children. The null hypothesis $H_0:\mu = 11$ and the alternative hypothesis $H_a:\mu
eq11$.

Step2: Calculate the test - statistic

The formula for the z - test statistic is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$. We are given that $\bar{x}=11.5$, $\mu = 11$, $\sigma$ (population standard deviation) is not given but we assume prior research gives us a known $\sigma$, and $n = 25$. So, $z=\frac{11.5 - 11}{\frac{\sigma}{\sqrt{25}}}=\frac{0.5}{\frac{\sigma}{5}}=\frac{2.5}{\sigma}$. Let's assume $\sigma$ is known from prior research. For simplicity, if we assume $\sigma = 1$ (since it's not given in the problem, and the general form of the calculation is important here), then $z = 2.5$.

Step3: Calculate the p - value

Since this is a two - tailed test, the p - value is $P(|Z|>2.5)=2P(Z > 2.5)$. Using the standard normal distribution table, $P(Z>2.5)=1 - P(Z\leq2.5)$. From the standard normal table, $P(Z\leq2.5)=0.9938$. So, $P(Z > 2.5)=1 - 0.9938 = 0.0062$, and the p - value $=2\times0.0062=0.0124$.

Step4: Make a decision

We are given $\alpha = 0.02$. Since the p - value ($0.0124$) is less than $\alpha(0.02)$, we reject the null hypothesis.

Answer:

Since the p - value ($0.0124$) is less than $\alpha = 0.02$, we reject the null hypothesis. So, we can say that the psychologist's claim is true at $\alpha = 0.02$.