Question

The time married men with children spend on child care averages 6.4 hours per week {Time, March 12, 2012). You belong lo 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 40 married couples will be used with the data collected showing the hours per week the husband spends on child care.

Use the sample data to perform a t-test to determine if the population mean number of hours married men are spending in child care differs from the mean reported by Time in your area? Use α = 0.05 as the level of significance.

1.

Hours
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
8.0
9.4
4.5
7.5
7.9
5.8
7.6
8.3
9.8
9.0
6.2
4.7
0.6
11.2
8.3
9.0
7.9
7.0
6.3
3.8
8.1
7.0
7.6
2.3
7.0

1. Whats the null and alternative hypothesis, what type of test is it?

2. Compute,

n

df

mean

std dev

std err

critical value

test value

p-value

3. Do you reject or fail to reject the null hypothesis. Explain why. Use the α = 0.05 level of significance.   Explain your conclusion using both the p-value and the test value.

4. State the conclusion?

Answer 1

(1)

Ho: = 6.4

Ha: 6.4

Null hypothesis states that the time married men with children spend on child care averages 6.4 hours per week.

Alternate hypothesis staets that the time married men with children spend on child care does not average 6.4 hrs. per week.

Hypothesis state to be conducted: One sample t test for mean

(2)

n = 40

df = 40 -1 = 39

mean = sum of all terms / no of terms = 280/ 40 = 7

standard deviation = s

data	data-mean	(data - mean)2
7.3	0.3	0.09
6.5	-0.5	0.25
7.7	0.7	0.49
5.8	-1.2	1.44
3.6	-3.4	11.56
9.7	2.7	7.29
6.2	-0.8	0.64
8.7	1.7	2.89
11.4	4.4	19.36
0.6	-6.4	40.96
8.2	1.2	1.44
7.2	0.2	0.04
9.4	2.4	5.76
7.5	0.5	0.25
5.4	-1.6	2.56
8	1	1
9.4	2.4	5.76
4.5	-2.5	6.25
7.5	0.5	0.25
7.9	0.9	0.81
5.8	-1.2	1.44
7.6	0.6	0.36
8.3	1.3	1.69
9.8	2.8	7.84
9	2	4
6.2	-0.8	0.64
4.7	-2.3	5.29
0.6	-6.4	40.96
11.2	4.2	17.64
8.3	1.3	1.69
9	2	4
7.9	0.9	0.81
7	0	0
6.3	-0.7	0.49
3.8	-3.2	10.24
8.1	1.1	1.21
7	0	0
7.6	0.6	0.36
2.3	-4.7	22.09
7	0	0

s = 2.4276

Standard error = SE

SE = 0.3838

t critical at df = 39, level of significance = 0.05 is +/- 2.02269092

t = 1.536

P value = TDIST (t statistics, df, 2) = TDIST(1.536, 39, 2) = 0.1261

p value = 0.1261

3.

As t stat (1.536) does not fall in the rejection area, we fail to reject the Null hypothesis.

Also as the p value (0.1261) is greater than level of significance (0.05), we fail to reject the Null hypothesis.

4. As we fail to reject the Null hypothesis at significance level 0.05, we do not have sufficient evidence to believe that the average time married men with children spend on child care is not euqal to 6.4 hours per week.