Question

From a random sample of 87 us adults with no more than a high school education that mean weekly income is 678 with a sample standard deviation of 197 from another random sample of 73 us adults with no more than a bachelor's degree the mean weekly income is 1837 with a standard deviation of $328 construct pain 95% confidence interval for the mean difference in the weekly income levels between us adults with no more than a high-school diploma and those with no more than a bachelor's degree there are a hundred 13 degrees of freedom in the appropriate probability distribution

Answer 1

So on this case the 95% confidence interval would be given by
`-1235.756 \leq \mu_1 -\mu_2 \leq -1072.245`. So we can conclude that we have a significant difference, with the mean of population two higher than the mean for the population 1, because the interval just contains negative values.

Explanation:

Notation and previous concepts

`n_1 =87` represent the sample of us adults with no more than a high school education

`n_2 =73` represent the sample of us adults with no more than a bachelor's degree

`\bar x_1 =678` represent the mean sample of us adults with no more than a high school education

`\bar x_2 =1837` represent the mean sample of us adults with no more than a bachelor's degree

`s_1 =197` represent the sample deviation of us adults with no more than a high school education

`s_2 =328` represent the sample deviation of us adults with no more than a bachelor's degree

`\alpha=0.05` represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:

`(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{((s^2_1)/(n_s)+(s^2_2)/(n_s))}` (1)

The point of estimate for
`\mu_1 -\mu_2` is just given by:

`\bar X_1 -\bar X_2 =678-1837=-1159`

The appropiate degrees of freedom are
`df=n_1+ n_2 -2=158`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,158)".And we see that
`t_(\alpha/2)=1.98`

The standard error is given by the following formula:

`SE=\sqrt{((s^2_1)/(n_s)+(s^2_2)/(n_s))}`

And replacing we have:

`SE=\sqrt{((197^2)/(87)+(328^2)/(73))}=43.816`

Confidence interval

Now we have everything in order to replace into formula (1):

`-1159-1.98\sqrt{((197^2)/(87)+(328^2)/(73))}=-1235.756`

`-1159+1.98\sqrt{((197^2)/(87)+(328^2)/(73))}=-1072.245`

So on this case the 95% confidence interval would be given by
`-1235.756 \leq \mu_1 -\mu_2 \leq -1072.245`. So we can conclude that we have a significant difference with the mean of population two higher than the mean for the population 1, because the interval contains just negative values.