Using the t-distribution, it is found that:
a) The 95% confidence interval for the population mean test score is (77.43, 80.57).
b) The margin of error is of 1.57.
What is a t-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 50 - 1 = 49 df, is t = 2.0096.
The other parameters are given by:
[tex]\overline{x} = 79, s = 5.4, n = 50[/tex]
Then, the bounds of the interval are given as follows.
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 79 - 2.0096\frac{5.4}{\sqrt{50}} = 77.43[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 79 + 2.0096\frac{5.4}{\sqrt{50}} = 80.57[/tex]
The 95% confidence interval for the population mean test score is (77.43, 80.57).
Item b:
The margin of error is half the difference between the bounds, hence:
M = (80.57 - 77.43)/2 = 1.57.
More can be learned about the t-distribution at https://brainly.com/question/16162795
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