Question

The following data represent a random sample of 9 marks (out of 10 ) on a statistics quiz. The marks are normally distributed with a standard deviation of 2 . Estimate the population mean with 90 % confidence.

Answer 1

We need to estimate the population mean with 90% confidence, for the given data of 9 marks on the statistics quiz, where the marks are normally distributed with a standard deviation of 2.

Sample of 9 marks are shown in the following table:

7

9

7

5

4

8

3

10

9

We are given these values as the population standard deviation \(\sigma=2\), and sample size \(n=9 .\)

The sample mean is calculated as shown below:

$$ \begin{aligned} \bar{x} &=\frac{\sum x_{i}}{n} \\ &=\frac{7+9+7+5+4+8+3+10+9}{9} \\ &=\frac{62}{9} \\ &=6.8889 \end{aligned} $$

The \(90 \%\) confidence interval estimator for the population mean is calculated as follows:

$$ C I=\bar{x} \pm z_{\alpha / 2} \frac{\sigma}{\sqrt{n}} $$

Where,

The sample mean is \(\bar{x}\),

The population standard deviation is \(\sigma\),

The size of the sample is \(n\),

$$ \begin{aligned} \bar{x} \pm z_{\alpha / 2} \frac{\sigma}{\sqrt{n}} &=\bar{x} \pm 1.645 \frac{\sigma}{\sqrt{n}} \\ &=6.8889 \pm 1.645 \frac{2}{\sqrt{2}} \\ &=\left(6.8889-1.645 \frac{2}{\sqrt{9}}, 6.8889+1.645 \frac{2}{\sqrt{9}}\right) \\ &=(5.7922,7.9856) \end{aligned} $$

Therefore, with \(90 \%\) confidence we can say that the population mean will be in the interval

$$ (5.7922,7.9856) . $$