The Basic Practice of Statistics
The Basic Practice of Statistics
8th Edition
ISBN: 9781319057916
Author: Moore
Publisher: MAC HIGHER
Question
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Chapter 7, Problem 7.59SE

a.

To determine

To make: The stem plot for IQ test scores of 31-seventh grade girls.

To check: Whether there are major departures from normality or not.

a.

Expert Solution
Check Mark

Answer to Problem 7.59SE

The stem plot is,

The Basic Practice of Statistics, Chapter 7, Problem 7.59SE , additional homework tip  1

No, there are no major departures from normality.

Explanation of Solution

Given info:

The data shows the IQ test scores of 31 seventh-grade girls in a Midwest school district.

Calculation:

Stemplot:

Software procedure:

Step-by-step software procedure to draw stemplot using MINITAB software is as follows:

  • Select Graph > Stem and leaf.
  • Select the column of IQ Test Scores in Graph variables.
  • Select OK.

The 1.5×IQR rule for outliers:

A observation is a suspected outlier, if it is more than Q3+(1.5×IQR) or less than Q1(1.5×IQR) .

Interquartile range (IQR):

The difference between the first quartile and third quartile is considered as interquartile range.

That is,

IQR=Q3Q1

Software procedure:

Step-by-step software procedure for the first quartile and third quartile in MINITAB software is as follows:

  • Choose Stat > Basic Statistics > Display Descriptive Statistics.
  • In Variables enter the columns IQ Test Scores.
  • Choose option statistics, and select first quartile, third quartile.
  • Click OK.

Output using MINITAB software is as follows:

The Basic Practice of Statistics, Chapter 7, Problem 7.59SE , additional homework tip  2

From Minitab output, the first quartile is 98.00 and third quartile is 114.00.

Substitute 114.00 for Q3 and 98.00 for Q1 in interquartile range

The interquartile range is,

IQR=114.0098.00=16

Substitute IQR in the 1.5×IQR rule

Q3+(1.5×IQR)=114.00+(1.5×16)=114.00+24=138

Q1(1.5×IQR)=98.00(1.5×16)=98.0024=74

The 1.5×IQR rule suspects one outlier in the IQ test scores of 31 seventh-grade girls because there is a value which is not in between limits Q1(1.5×IQR)=74 and Q3+(1.5×IQR)=138 .

Justification:

The stemplot showing that there are two low IQ’s. They are 72 and 74but for the small data, the IQ test scores of 31 seventh-grade girls is approximately normal.

b.

To determine

To obtain: The mean x¯ and standard deviation s and the proportions of the scores are within 1 standard deviation and within 2 standard deviations of the mean.

To check: Whether the proportions show an exactly normal distribution.

b.

Expert Solution
Check Mark

Answer to Problem 7.59SE

The mean x¯ is 105.84 and standard deviation s is 14.27.

The proportions of the scores are within 1 standard deviation of the mean, which is 0.742.

The proportions of the scores are within 2 standard deviation of the mean, which is 0.935.

If the distribution is exactly normal, then its proportion should be about 68% of the observations fall within σ of the mean μ and about 95% of the observations fall should be within 2σ of the mean μ .

Explanation of Solution

68-95-99.7 Rule:

If the proportions be in an exactly normal distribution then it is,

About 68% of the observations fall within σ of the mean μ .

About 95% of the observations fall within 2σ of the mean μ .

About 99.7% of the observations fall within 3σ of the mean μ .

Calculation:

For Mean and Standard deviation:

Software procedure:

Step-by-step software procedure for mean and standard deviation in MINITAB software is as follows:

  • Choose Stat > Basic Statistics > Display Descriptive Statistics.
  • In Variables enter the columns IQ Test Scores.
  • Choose option statistics, and select mean, standard deviation.
  • Click OK.

Output using MINITAB software is as follows:

The Basic Practice of Statistics, Chapter 7, Problem 7.59SE , additional homework tip  3

From Minitab output, the mean is 105.84 and standard deviation is 14.27.

The scores are within1 standard deviation and within 2 standard deviations of the mean that are obtained by finding x±1s and x±2s .

Substitute 105.84 for x and 14.27 for s,

x+1s=105.84+1×14.27x+1s=120.11

x1s=105.841×14.27x1s=91.57

x+2s=105.84+2×14.27x+2s=134.38

x2s=105.842×14.27x2s=77.3

For proportion of scores within 1 standard deviation:

The IQ test scores in between [91.57,120.11] is 23 and the total number IQ test scores is 31.

The formula to find the proportion of scores within 1 standard deviation is,

Proportion of IQ scores = Number of IQ test scores in between [91.57,120.11]Total number of IQ scores

Substitute 23 for ‘Number of IQ test scores in between [91.57,120.11] and 31 for ‘Total number of IQ scores’.

Proportion of IQ scores =2331=0.74190.742

Thus, the proportion of scores within 1 standard deviation is 0.742.

For proportion of scores within 2 standard deviations:

The IQ test scores in between [77.3,134.38] is 29 and the total number IQ test scores is 31.

The formula to find the proportion of scores within 2 standard deviations is,

Proportion of IQ scores = Number of IQ test scores in between [77.3,134.38]Total number of IQ scores

Substitute 29 for ‘Number of IQ test scores in between [77.3,134.38] and 31 for ‘Total number of IQ scores’.

Proportion of IQ scores =2931=0.935

Thus, the proportion of scores within 2 standard deviations is 0.935.

Justification:

If the distribution is exactly normal, then it should satisfy about 68% of the observations fall within σ of the mean μ and about 95% of the observations fall within 2σ of the mean μ .

Here, the proportion of scores within 1 standard deviation is 74.2% and the proportion of scores within 2 standard deviations is 93.5%. Thus, the proportions of scores are reasonably close to the proportion of exactly normal distribution.

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