Chapter 8.2, Problem 38E

a.

To determine

To find the confidence interval for the mean amount of sleeps from the remaining values. (not including the outlier)

Answer to Problem 38E

CI = $\left(6.66,8.03\right)$

Explanation of Solution

Given information :

Data points:

8.0 7.5 8.5 6.5 8.5 6.5 7.0 7.5

Calculation:

The outlier is value which is far away from the mean value or it is the value which lies in most extreme tail of the distribution.

Since, sample size 7 (n<30) and population standard deviation is not known therefore, here t-distribution is applicable.

Sample mean and standard deviation from the sample data can be computed as:

  $\begin{array}{l}\overline{X}=\frac{{\displaystyle \sum _{i\in N}{x}_i}}{n}\\ =\frac{ 7.5+8.0+6.5+8.5+6.5+7.0+7.5 }{7}\\ =7.35\end{array}$

Standard deviation;

  $\begin{array}{c}s=\sqrt{\frac{{\displaystyle \sum _{i=1}^{12}{\left(x-\mu \right)}^2}}{n-1}}\\ =\sqrt{\frac{2\times {\left(7.5-7.35\right)}^2+{\left(8-7.35\right)}^2+2\times {\left(6.5-7.35\right)}^2+{\left(8.5-7.35\right)}^2+{\left(7-7.35\right)}^2}{7-1}}\\ =0.748\end{array}$

The 95% confidence interval for the mean amount of sleep.

  $\overline{x}\pm t_{\frac{\alpha }{2},n-1}\times \frac{s}{\sqrt{n}}$

Degree of freedom = 7-1 = 6.

Thus, t- critical (table) value at 5% significance level and 6 degree of freedom can be seen from table.

i.e. t =2.44

  $\begin{array}{c}CI=\overline{x}\pm t_{\frac{\alpha }{2},n-1}\times \frac{s}{\sqrt{n}}\\ =7.35\pm 2.44\times \frac{0.748}{\sqrt{7}}\\ =\left(6.66,8.03\right)\end{array}$

b.

To determine

To find the confidence interval for the mean amount of sleep when outlier in the data.

Answer to Problem 38E

Yes, the results are different.

Checking for outlier gives the good idea about data whether it is normally distributed or not.

Explanation of Solution

To check the 95% confidence interval for the mean amount of sleep when outlier is present.

Sample mean and standard deviation in presence of outlier can be computed as:

  $\begin{array}{l}\overline{X}=\frac{{\displaystyle \sum _{i\in N}{x}_i}}{n}\\ =\frac{ 7.5+8.0+6.5+8.5+6.5+7.0+7.5 +24}{8}\\ =9.43\end{array}$

Standard deviation:

  $\begin{array}{c}s=\sqrt{\frac{{\displaystyle \sum _{i=1}^{12}{\left(x-\mu \right)}^2}}{n-1}}\\ =\sqrt{\frac{2\times {\left(7.5-7.35\right)}^2+{\left(8-7.35\right)}^2+2\times {\left(6.5-7.35\right)}^2+{\left(8.5-7.35\right)}^2+{\left(7-7.35\right)}^2+{\left(24-7.35\right)}^2}{8-1}}\\ =6.33\end{array}$

The 95% confidence interval for the mean amount of sleep.

  $\overline{x}\pm t_{\frac{\alpha }{2},n-1}\times \frac{s}{\sqrt{n}}$

Degree of freedom = 8-1 = 7.

Thus, t- critical (table) value at 5% significance level and 6 degree of freedom can be seen from table.

i.e. t =2.36

  $\begin{array}{c}CI=\overline{x}\pm t_{\frac{\alpha }{2},n-1}\times \frac{s}{\sqrt{n}}\\ =9.43\pm 2.36\times \frac{6.33}{8}\\ =\left(7.56,11.29\right)\end{array}$

Since, it can be easily seen that after including outlier the size of or width of confidence interval.

Hence, it can be concluded that hypothesize mean of quality control manager does not lie in CI.

Therefore, it is not reasonable to considered that mean.

Since, outlier is point which is far away from the other data points or mean value and it affects all descriptive parameters and normality of the distribution as well.

So, it is necessary to check the outlier in the data set