Chapter 8.2, Problem 37E

Eat your kale: Kale is a type of cabbage commonly found in salad and used in cooking in many parts of the world. Six measurements were made of the mineral content (in percent) of kale: with the following results-

26.1 17.5 15.4 164 15.1 12.8

It turns out that the value 26.1 came from a specimen that the investigator forgot to wash before measuring.

1. The data contain an outlier that is clearly a mistake. Eliminate the outlier, then construct a 95% confidence interval for the mean mineral content from the remaining values.
2. Leave the outlier in and construct the 95% confidence interval. Are the results noticeably different? Explain why it is important to check data for outliers.

To determine

To find the confidence interval for the mean mineral content from the remaining values. (not including the outlier value)

Answer to Problem 37E

  $CI=\left(13.27,17.60\right)$

Explanation of Solution

Given information :

Data points:

26.1 17.5 15.4 16.4 15.1 12.8

Calculation:

x	$\left(x-\overline{x}\right)$	${\left(x-\overline{x}\right)}^2$
17.5	2.06	4.2436
15.4	-0.04	0.0016
16.4	0.96	0.9216
15.1	-0.34	0.1156
12.8	-2.64	6.9696
${\displaystyle \sum x=77.2}$		${\displaystyle \sum {\left(x-\overline{x}\right)}^2}=12.252$

Mean:

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{17.5+15.4+\mathrm{............}+12.8}{5}\\ =15.44\end{array}$

Sample standard deviation:

  $\begin{array}{c}{\sigma }_{\overline{x}}=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{12.252}{5-1}}\\ =1.750\end{array}$

95% confidence interval can be computed as:

Since, sample size is less than 30 and population standard deviation is not known so , t-test will be used.

Degree of freedom = 5-1

  = 4

t-value = 2.77

  $\begin{array}{l}\overline{x}\pm t_{\alpha |2,n-1}\times \frac{{\sigma }_{\overline{x}}}{\sqrt{n}}\\ 15.44\pm 2.77\times \frac{1.750}{\sqrt{5}}\\ \left(13.27,17.60\right)\end{array}$

To determine

To check the claim Whether online course and traditional course are same.

Answer to Problem 37E

CI = $\left(12.36,22.05\right)$ wider as compare to part a)

Explanation of Solution

Calculation:

x	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
26.1	8.89	79.0321
17.5	0.29	0.0841
15.4	-1.81	3.2761
16.4	-0.81	0.6561
15.1	-2.11	4.4521
12.8	-4.41	19.4481
${\displaystyle \sum x}=103.3$	$$	${\displaystyle \sum {\left(x-\overline{x}\right)}^2}=106.948$

Mean:

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{17.5+15.4+\mathrm{............}+12.8}{6}\\ =17.21\end{array}$

Sample standard deviation:

  $\begin{array}{c}{\sigma }_{\overline{x}}=\sqrt{\frac{{\displaystyle \sum {\left(X-\overline{X}\right)}^2}}{N-1}}\\ =\frac{106.948}{6-1}\\ =4.62\end{array}$

95% confidence interval can be computed as:

Since, sample size is less than 30 and population standard deviation is not known so, t-test will be used.

Degree of freedom = 6-1

  = 5

t-critical value = 2.57 (t-value is taken from t-table)

  $\begin{array}{l}\overline{x}\pm t_{\alpha |2,n-1}\times \frac{{\sigma }_{\overline{x}}}{\sqrt{n}}\\ 17.21\pm 2.57\times \frac{4.62}{\sqrt{6}}\\ \left(12.36,22.05\right)\end{array}$

As compare to part a) the confidence interval width has increased since, it is more likely to contain the true population 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