Question

In studying the dairy food expenditure of families in a city, random samples of fifty families and forty families are selected from urban area and rural area, respectively. Selected families from urban area have a mean of $150 and a standard deviation of $30. Selected families from rural area have a mean of $135 and a standard deviation of $25. Construct a 95% confidence interval for the difference of mean daily food expenditure between all families of urban area and rural area. State your assumptions or approximations.

Answer 1

1 represents the urban area and 2 represents the rural area.

We need to construct the 95% confidence interval for the difference between the population means μ1​−μ2​, for the case that the population standard deviations are not known.

Sample Mean 1 =	150
Sample Standard Deviation 1 =	30
Sample Size 1 =	50
Sample Mean 2 =	135
Sample Standard Deviation 2 =	25
Sample Size 2 =	40

We assume that the population variances are equal, so then the number of degrees of freedom are

df=n1​+n2​−2=50+40−2=88.

The critical value for α=0.05 and df = 88 degrees of freedom is .

The corresponding confidence interval is computed as shown below:

Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows:

Since we assume that the population variances are equal, the standard error is computed as follows:

Now, we finally compute the confidence interval:

Therefore, the 95% confidence interval for the difference between the population means 3.24<μ1​−μ2​<26.76, which indicates that we are 95% confident that the true difference between population means is contained by the interval (3.24, 26.76)(3.24,26.76).