In a sample of 105 married heterosexual couples, the average age difference (husband's age - wife's age) was 2.829 years with a standard deviation of 4.995 years. These summary statistics were taken from a data set from the Lock5 textbook. Is there evidence that, on average, in the population, husbands tend to be older than their wives?
First we need to check our assumptions. In this case the sample size is greater than 30 so we can use the t-distribution.
We know n = 105, \(\mu_{\text{husband's age}}-\mu_{\text{wife's age}}=2.829\), and \(s=4.995\). Since we want to know if the husbands are older than their wives then our difference in ages would be positive. So our alternative hypothesis is \(H_a\colon \gt 0\).
To complete this using Minitab...
- Select Stat > Basic Statistics > Paired t
- Select Summarized data (differences)
- For sample size enter 105, enter 2.829 for sample mean and 4.995 for standard deviation.
- Select Options
- The Hypothesized difference should be 0. (or 0.0)
- Select Difference > hypothesized difference for the Alternative hypothesis
- Click OK and OK
Estimation for Paired Difference
Mean | StDev | SE Mean | 95% CI for \(\mu_d\) |
---|---|---|---|
105 | 2.829 | 4.995 | (1.862, 3.796 |
\(\mu\)_difference: population mean of (Sample 1 - Sample 2)
Test
Null hypothesis | H0: \(\mu\)_difference = 0 |
---|---|
Alternative hypothesis | H1: \(\mu\)_difference ≠ 0 |
T-Value | P-Value |
---|---|
5.80 | 0.000 |
Interpret the results
Because the sample size is large (\(n \ge 30\)), the t distribution may be used to approximate the sampling distribution.
\(H_{0}:\mu_d=0\)
\(H_{a}:\mu_d \gt 0\)
Null hypothesis | H0: \(\mu_d\) = 0 |
---|---|
Alternative hypothesis | H1: \(\mu_d\) > 0 |
T-Value | P-Value |
---|---|
5.80 | 0.000 |
The t test statistic is 5.80.
From the output, the p value is 0.000
\(p\leq .05\), therefore our decision is to reject the null hypothesis
There is evidence that in the population, on average, the husband's age in heterosexual couples is greater than the wife's age.