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Statistics / Fundamentals of Statistics 4 / Chapter 3.2 / Problem 39AYU

Fundamentals of Statistics | 4th Edition | ISBN: 9780321838704 | Authors: Michael Sullivan,III

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Solution: Resistance and Sample Size Each of the following

Chapter 9, Problem 39AYU

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QUESTION:

Problem 39AYU

Problem

Resistance and Sample Size Each of the following three data sets represents the IQ scores of a random sample of adults. IQ scores are known to have a mean and median of 100. For each data set, determine the sample standard deviation. Then recompute the sample standard deviation assuming that the individual whose IQ is 106 is accidentally recorded as 160. For each sample size, state what happens to the standard deviation. Comment on the role that the number of observations plays in resistance.

SAMPLE OF SIZE 5
106	92	98	103	100

SAMPLE OF SIZE 12
106	92	98	103
98	124	83	70

SAMPLE OF SIZE 30
106	92	98	103	100	102
98	124	83	70	108	121
102	87	121	107	97	114
140	93	130	72	81	90
103	97	89	98	88	103

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QUESTION:

Problem 39AYU

Problem

Resistance and Sample Size Each of the following three data sets represents the IQ scores of a random sample of adults. IQ scores are known to have a mean and median of 100. For each data set, determine the sample standard deviation. Then recompute the sample standard deviation assuming that the individual whose IQ is 106 is accidentally recorded as 160. For each sample size, state what happens to the standard deviation. Comment on the role that the number of observations plays in resistance.

SAMPLE OF SIZE 5
106	92	98	103	100

SAMPLE OF SIZE 12
106	92	98	103
98	124	83	70

SAMPLE OF SIZE 30
106	92	98	103	100	102
98	124	83	70	108	121
102	87	121	107	97	114
140	93	130	72	81	90
103	97	89	98	88	103

ANSWER:

Answer :

Step 1

Given the 5 sample size.

Sample size : 106,92,98,103,100

First we need to find the sample variance.

mean = $\frac{\sum\limits_{i=1}^{n}{x}_{i}}{n}$

mean = $\frac{499}{5}$

mean = 99.80

Then the standard deviation is

${s}^{2}$ = $\frac{\Sigma{}{\left({{x}_{i}-\overline{x}}\right)}^{2}}{n-1}$

${s}^{2}$ = $\frac{38.44+60.84+10.24+0.04}{5-1}$

${s}^{2}$ = $\frac{112.8}{4}$

${s}^{2}$ = 28.2

Then the sample variance is

s= $\sqrt{\ \frac{\Sigma{}{\left({{x}_{i}-\overline{x}}\right)}^{2}}{n-1}\ \ }$

s = $\sqrt{\ 28.2\ }$

s = 5.3

Hence the sample variance is s = 5.3

Then recompute the sample standard deviation assuming that the individual whose IQ is 106 is accidentally recorded as 160.

Given the 5 sample size.

Sample size : 160,92,98,103,100

mean = $\frac{\sum\limits_{i=1}^{n}{x}_{i}}{n}$

mean = $\frac{553}{5}$

mean = 110.60

Then the standard deviation is

${s}^{2}$ = $\frac{\Sigma{}{\left({{x}_{i}-\overline{x}}\right)}^{2}}{n-1}$

${s}^{2}$ = $\frac{2440.36+345.96+158.76+57.76+112.36}{5-1}$

${s}^{2}$ = $\frac{3115.2}{4}$

${s}^{2}$ = 778.800

Then the sample variance is

s= $\sqrt{\ \frac{\Sigma{}{\left({{x}_{i}-\overline{x}}\right)}^{2}}{n-1}\ \ }$

s = $\sqrt{\ 778.800}$

s = 27.9

Incorrect s = 27.9.

b)

Here the sample size is n=12

Then the standard deviation is

${s}^{2}$ = $\frac{\Sigma{}{\left({{x}_{i}-\overline{x}}\right)}^{2}}{n-1}$

${s}^{2}$ = $\frac{2376.99}{11}$

${s}^{2}$ = 216.09

Then the sample variance is

s= $\sqrt{\ \frac{\Sigma{}{\left($

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