Question

The time married men with children spend on child care averages 9.8 hours per week (Time, March 12, 2012). You belong to a professional group on family practices that would like to do its own study to determine if the time married men in your area spend on child care per week differs from the reported mean of 9.8 hours per week. The standard deviation for the amount of time married men in your area spend on child care is believed to be a constant 3.9 hours per week. A sample of 200 couples produced a mean of 9.1 hours spents on child care per week.

Using the .02 level of significance, test whether the population mean number of hours married men are spending on child care differs from the mean of 9.8 hours reported by Time magazine.

Round your critical value to 2 decimal places and your p value 4 decimal places

Answer 1

Solution:

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H₀: The population mean number of hours married men are spending on child care is 9.8 hours

Alternative hypothesis: H_a: The population mean number of hours married men are spending on child care differs from the mean of 9.8 hours.

H₀: µ = 9.8 versus H_a: µ ≠ 9.8

This is a two tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 9.8

Xbar = 9.1

σ = 3.9

n = 200

α = 0.02

Critical value = - 2.33 and 2.33

(by using z-table or excel)

Z = (9.1 – 9.8)/[3.9/sqrt(200)]

Z = -2.5383

P-value = 0.0111

(by using Z-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the population mean number of hours married men are spending on child care differs from the mean of 9.8 hours.