Question

A random sample of 28 statistics tutorials was selected from the past 5years and the percentage of students absent from each one was recorded. The results are given below. Assume the percentage of student's absences are approximately normally distributed. Use Excel to estimate the mean percentage of absences per tutorial over the past 5 years with 90% confidence.

Answer 1

Using the t-distribution, the 90% confidence interval for the mean number of absences is given by: (9.22, 11.60).

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 28 - 1 = 27 df, is t = 1.7033.

Researching this problem on the internet, the parameters are given by:

$\overline{x} = 10.41, s = 3.71, n = 28$.

Hence the bounds of the interval are given by:

• $\overline{x} - t\frac{s}{\sqrt{n}} = 10.41 - 1.7033\frac{3.71}{\sqrt{28}} = 9.22$

• $\overline{x} + t\frac{s}{\sqrt{n}} = 10.41 + 1.7033\frac{3.71}{\sqrt{28}} = 11.60$

The 90% confidence interval for the mean number of absences is given by: (9.22, 11.60).

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

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