The time married men with children spend on child care averages 6.5 hours per week.

Question

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 6.5 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. The sample data is below.

6.8| 9.4| 7.5| 5.4| 9.4| 7.6| 7.8| 7.3| 3.7| 7.7| 7.2| 8.2| 0.8| 7.2| 9.0| 7.6| 7.2| 9.9| 6.5| 11.7| 6.4| 7.5| 10.4| 7.4| 9.9 8.0| 6.4| 7.8| 8.5| 4.4| 3.4| 8.4| 2.4| 4.6| 5.9| 0.8| 5.9| 12.0| 8.5| 10.3

1. What is the sample mean (to 2 decimals)?
2. Calculate the value of the test statistic (to 2 decimals).
3. What is the p-value?

Answer 1

Hi Vi,

We do not know population standard deviation here, so we have to use a t-test statistic. Keep that in mind. With that in mind, I'll break the question into the three parts as a, b, and c:

(a) Add up all values you were given in sample data and divide by 40. Use a calculator since this is cumbersome manually. I got: x-bar=7.17, x-bar is the notation for sample mean.

(b) Remember, we are using t because we do not have population standard deviation. Formula for a t-test statistic is:

t=(x-bar-mu/SE)

Let's break this down:

t=test statistic, ultimately what we're after for this part

x-bar=sample mean

mu=proposed population mean

SE=standard error

Now, standard error has its own formula and calculation. It is:

SE=s/sqrt(n) where:

s=sample standard deviation

n=sample size

We have the sample size, but we need to compute sample standard deviation. I recommend an online calculator, a manual graphics calculator, or statistical software. Otherwise, we're looking at subtracting and squaring 40 measurements! How about no! From online calculator:

s=2.55

SE=s/sqrt(n)

s=2.55

n=40

SE=2.55/sqrt(40)

SE=0.40

Now, let's go back to our t-statistic formula:

t=(x-bar-mu)/SE

x-bar=7.17, calculated in part (a)

mu=6.5, given

SE=0.40

t=(7.17-6.5)/0.40

t=1.68

(c) This is a little tricky. To get an exact p-value, you need statistical software, and the program you use varies by school. Manually, we can only get an approximation, which I will explain here:

First of all, we need to know degrees of freedom. For one sample t, this is:

df=n-1

df=40-1

df=39

Now, we already have our t-test statistic 1.68, so we go to the t-table. We want to be closest to 39 degrees of freedom without going over. On my table, this means we have to go with 30 degrees of freedom, so search that row for 1.68. It doesn't come up, but we do have boundary values of 1.310 and 1.697. Now, look at the alpha levels at the top of the table for two-sided at both of these boundary marks. We were not told if we were looking for lesser or greater, so we have to assume both. We can conclude:

0.10<p<0.20

True p-value likely leans toward the lower end around 0.11 or 0.12, but we cannot safely be more precise than that without statistical software. I hope this helps.