Construct a 95% confidence interval for the population mean, μ. Assume the population has a normal distribution. A sample of 25 randomly selected students has a mean test score of 81.5 with a standard deviation of 10.2.

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Answer:

The 95% confidence interval for the population mean is between 77.5 and 85.5

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 1.96*\frac{10.2}{\sqrt{25}} = 4[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 81.5 - 4 = 77.5.

The upper end of the interval is the sample mean added to M. So it is 81.5 + 4 = 85.5

The 95% confidence interval for the population mean is between 77.5 and 85.5