Using the z-distribution, it is found that the 95% confidence interval for the difference is (-1.3, -0.7).
Considering the data given:
[tex]\mu_S = 3.8, n = 175, s_S = \frac{1.7}{\sqrt{175}} = 0.1285[/tex]
[tex]\mu_N = 4.8, n = 152, s_N = \frac{1.2}{\sqrt{152}} = 0.0973[/tex]
The mean is the subtraction of the means, hence:
[tex]\overline{x} = \mu_S - \mu_N = 3.8 - 4.8 = -1[/tex]
The standard error is the square root of the sum of the variances of each sample, hence:
[tex]s = \sqrt{s_S^2 + s_N^2} = \sqrt{0.1285^2 + 0.0973^2} = 0.1612[/tex]
It is given by:
[tex]\overline{x} \pm zs[/tex]
We have a 95% confidence interval, hence the critical value is of z = 1.96.
Then, the bounds of the interval are given as follows:
More can be learned about the z-distribution at https://brainly.com/question/25890103
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