Question

Six data sets are presented, some of them are samples from a normal distribution, and some of them are samples from populations that are not normally distributed. Identify the samples that are not from normally distributed populations.

L1: Drug concentration six hours after administration

L2: Reading scores on standardized test for elementary children

L3: The number of minutes clerical workers took to complete a certain worksheet

L4: The level of impurities in aluminum cans (in percent)

L5: The number of defective items produced during an hour

L6: The weight of trout in ounces

L1	L2	L3	L4	L5	L6
4.7	72	4.5	2.1	21	9.9
3.2	77	5.8	1.3	16	11.3
5.1	65	3.7	2.8	10	11.4
4.7	85	4.9	1.5	10	9
3.0	68	4.3	1.0	11	10.1
3.4	83	4.7	8.2	9	8.2
4.4	73	5.8	1.9	13	8.9
3.5	79	3.2	9.5	12	9.9
4.5	72	3.0	3.2	11	10.5
5.8	81	5.1	1.3	29	8.6
3.7	79	3.6	4.4	10	7.8
4.9	91	4.3	3.8	14	10.8
5.4	69	3.6	2.7	27	8.4
3.6	67	5.4	8.0	10	9.6
4.3	82	4.7	1.9	11	9.9
3.0	66	3.0	4.9	11	8.4
5.1	77	3.4	4.5	12	9.0
4.3	66	4.3	1.5	19	9.1

Look at your modified boxplots. Which boxplot(s) show(s) outliers in the data?

		Drug concentration six hours after administration
		Reading scores on standardized test for elementary children
		The number of minutes clerical workers took to complete a certain worksheet
		The level of impurities in aluminum cans (in percent)
		The number of defective items produced in an hour
		The weight of trout in ounces

Answer 1

To understand whether the sample is from normal population or not, we can use any noramlity tests like K-S test, Anderson Darling test, Shapiro-Wilk test, etc. Here we are using Shapiro-Wilk test. If the p-value is greater than 0.05, we consider the data to be normal.

Here, we can see that L1,L2,L3,L6 are normal, but L4 and L5 not normal. The boxplots of L4 and L5 show outliers.

Note: Attaching below test results and boxplots for each data.

> shapiro.test(L1)

        Shapiro-Wilk normality test

data:  L1
W = 0.9532, p-value = 0.4773

> shapiro.test(L2)

        Shapiro-Wilk normality test

data:  L2
W = 0.94656, p-value = 0.3736

> shapiro.test(L3)

        Shapiro-Wilk normality test

data:  L3
W = 0.94718, p-value = 0.3825

> shapiro.test(L4)

        Shapiro-Wilk normality test

data:  L4
W = 0.83787, p-value = 0.005484

> shapiro.test(L5)

        Shapiro-Wilk normality test

data:  L5
W = 0.76241, p-value = 0.0004688

> shapiro.test(L6)

        Shapiro-Wilk normality test

data:  L6
W = 0.95834, p-value = 0.5699