Question

The results in the form of a Minitab output are shown below. Use this output to answer the questions that follow.

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Descriptive Statistics: HOURS

Variable   N         Mean        SE Mean       StDev        Minimum     Q1        Median     Q3    Maximum

HOURS     95      16.6            0.318            3.10            6.0               15.0       16.0          18.0   24.0

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1. Measures of Central Tendency: Compare the Mean to the Median. What is your conclusion regarding the shape of the distribution, degree of skewness, and the possible presence of outliers? Explain your answer in the space provided below.

2. Measures of Variation: Find the value for the Mean and the Standard Deviation in the Minitab output.
   1. Use these statistics to calculate the Coefficient of Variation. Show your calculation and final answer in the space provided below.
   2. Then interpret the value of CV. How would you use this statistic to draw a conclusion regarding the degree of variability among values in the sample?

3. Standardized Values: Assume that a randomly selected consumer reported that she spends 26 hours per week online.
   1. Calculate the equivalent Z-score for this consumer’s weekly online activity. Show your calculation and final answer in the space provided below.
   2. Using your calculation of the converted Z-score, would you conclude that this consumer is an outlier? Why or why not?

4. Empirical Rule: Assume a near normal distribution exists for this population. Answer the following questions regarding the values found in the Minitab output for this study. Show your calculations and final answer in the spaces provided.

1. 68% of all observations will fall within two values. What are these two values?

2. 95% of all observations will fall within two values. What are these two values?

3. Virtually all (over 99%) of all observations will fall within two values. What are these two values?

Answer 1

1) Measures of central tendency: Mean is 16.6 and Median is 16.0, so mean is slightly (by 0.6) greater than median.

Since, Mean is approximately equal to median, we can say that shape of distribution is Normal. We can confirm the same by calculating Pearson's coefficient of skewness =[ 3 * (mean - median) ] / standard deviation.

Pearson's coefficient of skewness = [3*(16.6-16) ] / 3.10 = 0.58. Since, this value is positive, we can say that the data is positively skewed.

Outlier can affects the mean, but since, mean is not much different than the median there is not possible presence of outlier.

2) Measures of variation: Coefficient of variation = standard deviation / mean = 3.10 / 16.6 = 0.1867 = 19% So, we can say that level of dispersion around mean is 19%.

3) Standardized values: Z-score = (x-mean) / standard deviation , here x is 26.

So, Z-score=( 26 - 16.6) / 3.10 = 3.0322 .

Since, Z-score is greater than 3 by empirical rule we can say that the consumer is an outlier.

4) If, 68% of all observations will fall within two values then those two values are and

= 16.6 - 3.10 = 13.5 and = 19.7

If, 95% of all observations will fall within two values then those two values are and

= 16.6 - (2*3.10) = 10.4 and = 22.8

and If, 95% of all observations will fall within two values then those two values are and

= 16.6 - (3*3.10) = 7.3 and = 25.9