Question

The time married men with children spend on child care averages 6.4 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 6.4 hours per week. A sample of 50 married couples will be used with the data collected showing the hours per week the husband spends on child care. The sample data is below.

Hours Spent in Child Care
7.3
6.5
7.7
5.8
3.6
9.7
6.2
8.7
11.4
0.6
8.2
7.2
9.4
7.5
5.4
8
9.4
4.5
7.5
7.9
5.8
7.6
8.3
9.8
9
6.2
4.7
0.6
11.2
8.3
9
7.9
7
6.3
3.8
8.1
7
7.6
2.3
7
8
7.5
9
8.2
7.9
9
8
10
6
8.2
a. Formulate the hypotheses that can be used to determine whether the population mean -number of hours married mean are spending in child care differs from the mean reported by Time in your area.
b. What is the sample mean and the p-value?
c. At each significant level of 90%, 95% or 99%, what is your conclusion?

Answer 1

H0 : μ = 6.4

H0 : μ > 6.4

Sample mean = 7.24

Pvalue = 0.120

At each α - level ; 0.1, 0.05, 0.01 ; We fail to reject the null

Explanation:

From the sample data :

Mean, xbar, = The test statistic :

ΣX / n = 361.8 / 50 = 7.24

Standard deviation, s = √Σ(x - xbar)²/ n-1 = 2.263

The hypothesis :

H0 : μ = 6.4

H0 : μ > 6.4

The test statistic, Z ;

(xbar - μ) / s/√n

(7.24 - 6.4) / s/√n

0.84 / 2.263/√10

Z = 0.84 / 0.7156234

Z = 1.174

Using the Pvalue from Zscore calculator to Obtain the Pvalue ;

Pvalue = 0.120

Decision region :

Reject H0 ; if Pvalue < α

At 90% ; α = 0.1

Pvalue > α ; We fail to reject the Null

At 95% ; α = 0.05

Pvalue > α ; We fail to reject the Null

At 99% ; α = 0.01

Pvalue > α ; We fail to reject the Null