Question

I literally have no idea what I’m doing. Help!!! Please show work!!

Write an essay using Word and post it as an attachment to the discussion. Cover the following points. Choose a data set from Stat Disk using one of the following files from 13th Edition Elementary Statistics: Oscar Winner Freshman 15 Word Count Garbage Weights Passive and Active Smoke If the file contains more than one variable, choose a quantitative variable. Create a histogram using your chosen variable and paste it into your file. Using the information on page 25 in the textbook, discuss the shape, center, variability, and outliers. Calculate the descriptive statistics in Stat Disk and paste your results in your file. Based on whether there are outliers, would the mean or median be the best measure of center for the data set? Using the value of the standard deviation discuss the amount of variation in the data. Create a boxplot and paste your results in your file. If you choose Modified Boxplot, the outliers will be displayed as asterisks if there are any. Give the five-number summary and discuss the range of the data set and if it is skewed left or right based on the boxplot. (Look at the position of the median in the box.) Based on the descriptive statistics, is the data left or right skewed (look at the mean and median)?

Answer 1

The variable chosen is number of winner actresses. The following table shows the required data: S.No. Actresses S.No. S.No. Actresses S.No. Actresses 21 Actresses Actresses 35 33 Actresses 26 80 22 23 29 33 22 au- 54 49 28 29 30 31 32 17 18 30 29 58 39 61 19 20 40 60 41 The first part of the question requires to create the histogram of the data. The required histogram is created using Minitab. The following are the steps to create the histogram: Step 1: Go to graph select the option histogram>select the option simple → ok Step 2: Select the variable to graph variable → ok. The snapshot of the histogram is shown below:

Histogram of Actresses Frequency 1 30 40 60 70 50 Actresses According to the histogram, the shape, center, variability, and outliers are discussed in the following points: 1. The distribution of the data is right-skewed because the peak of the histogram is left of the center. Thus, for right-skewed distribution mode<median<mean. 2. The centers of the data are mean, median and mode. The central tendencies are obtained using Minitab as shown in the following steps: Step 1: Go to Stat>select basic statistics select basic descriptive statistics.

Step 2: Select the variable click on statistics option select the required display descriptive statistics → ok. The required centers of the data are shown in the following snapshot of the output: Descriptive Statistics: Actresses Statistics Variable Actresses N for Mean Median Mode Mode 36.24 33.00 29 8 Since the distribution of the data is right-skewed, therefore, the appropriate center of the data is median, that is, Median = 33 3. The spread in the histogram is noticeable, so the given data is highly variable. The exact value of the variability is quantified using the standard deviation of the data. The standard deviation is obtained using Minitab using the following steps: Step 1: Go to Stat+select basic statistics select basic descriptive statistics. Step 2: Select the variable click on statistics optionselect the required display descriptive statistic → ok. The required centers of the data are shown in the following snapshot of the output: Descriptive Statistics: Actresses Statistics Variable Actresses StDev 11.47

4. Since the bars are observed in the histogram at the extreme right with a break after the bin 60-70, therefore outliers are present in the data. The descriptive statistics are obtained using Minitab using the following steps: Step 1: Go to Stat→select basic statistics select basic descriptive statistics. Step 2: Select the variable> click on statistics option> select the display descriptive statistics → ok. The required centers of the data are shown in the following snapshot of the output:

Descriptive Statistics: Actresses Statistics Total Variable Count Mean SE Mean StDev Variance Coefvar Minimum Q1_Median Q3 Actresses 87 36.24 1.23 11.47 131.49 31.64 21.00 29.00 33.00 41.00 N for Variable Maximum Range IQR Mode Mode Skewness Kurtosis Actresses 80.00 59.00 12.00 29 8 1 .61 2.94 Since the outliers are observed in the data, therefore the appropriate measure of center is median because the median is not affected by outliers in the data set. The value of the standard deviation from the above snapshot of the output is o=11.47. Thus, the amount of variation in the data is 11.47, which implies that the average distance between the values of the data and its mean is 11.47. The boxplot is constructed using Minitab as shown using the following steps: Step 1: Go to graph>select the option boxplotselect the option simple → ok Step 2: Select the variable to graph variable select the options from the boxplot data view ok. The snapshot of the boxplot is shown below:

Boxplot of Actresses Actresses The five-number summary of the given data is obtained using the following steps: Step 1: Go to Stat select basic statistics select basic descriptive statistics. Step 2: Select the variable click on statistics option select minimum, first quartile, median, third quartile and maximum from display descriptive statistics → ok. The five-number summary of the data is shown in the following snapshot of the output:

Descriptive Statistics: Actresses Statistics Variable Actresses Minimum Q1 Median Q3 Maximum 21.00 29.00 33.00 41.00 80.00 The range of the data is obtained as shown below: Range = max - min = 80-21 = 59 Thus, the range of the data is 59 Since the boxplot is lopsided which implies that the longer side of the box plot lies to the right side of the median. Thus, the distribution of the data is right-skewed. Since, mean>median, therefore, the data is right-skewed based on descriptive statistics.