Question

You are told that in a sample of size 225 the mean is 48.5 and the standard deviation is 1.8. the study is reported with 90% confidence level. explain how to determine if 48.8 is within the confidence interval.

Answer 1

Using the t-distribution, it is found that the 90% confidence interval is (48.3, 48.7), hence 48.8 is not within the interval.

What is a t-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• t is the critical value.

• n is the sample size.

• s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 90% confidence interval, with 225 - 1 = 224 df, is t = 1.6517.

The other parameters are given as follows:

$\overline{x} = 48.5, s = 1.8, n = 225$.

Hence the bounds of the interval are given by:

$\overline{x} - t\frac{s}{\sqrt{n}} = 48.5 - 1.6517\frac{1.8}{\sqrt{225}} = 48.3$

$\overline{x} + t\frac{s}{\sqrt{n}} = 48.5 + 1.6517\frac{1.8}{\sqrt{225}} = 48.7$

The 90% confidence interval is (48.3, 48.7), hence 48.8 is not within the interval.

More can be learned about the t-distribution at brainly.com/question/16162795

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