Examples

BA501 : The Class : Stats : Intervals : Examples
Sampling Distributions: Examples
Examples

§ Sampling Distributions and Confidence Intervals §

Lesson 2: Examples

§ Central Limit Theorem §

§ Conversion of xbars to Z Scores §

§ Probability of x and xbar §

§ Confidence Interval using Z §

§ Confidence Interval using t §

§ Sample Size §

Example 1: Central Limit Theorem

Assume a normally distributed population of the life time of batteries with mean life time 250 minutes and standard deviation 25 minutes when used in an ordinary toy. Draw 20 samples (each sample size of n = 10, ten randomly selected batteries) from the of the population of batteries.

Sample 1: 249, 252, 245, 249, 241, 278, 239, 296, 257, 244

xbar₁ = 255.00, s₁ = 18.16.

Sample 2: xbar₂ = 239.6, s₂ = 18.09

Sample 3: xbar₃ = 249.9, s₃ = 21.50

Sample 4: xbar₄ = 235.6, s₄ = 29.82

Sample 5: xbar₅ = 263.8, s₅ = 23.13

Sample 6: xbar₆ = 253.7, s₆ = 19.55

Sample 7: xbar₇ = 250.3, s₇ = 24.95

Sample 8: xbar₈ = 252.5, s₈ = 29.46

Sample 9: xbar₉ = 248.7, s₉ = 23.70

Sample 10: xbar₁₀ = 246.4, s₁₀ = 33.90

Sample 11: xbar₁₁ = 249.5, s₁₁ = 36.80

Sample 12: xbar₁₂ = 263.2, s₁₂ = 21.67

Sample 13: xbar₁₃ = 247.3, s₁₃ = 26.44

Sample 14: xbar₁₄ = 255.1, s₁₄ = 28.30

Sample 15: xbar₁₅ = 251.2, s₁₅ = 25.02

Sample 16: xbar₁₆ = 254.7, s₁₆ = 25.48

Sample 17: xbar₁₇ = 243.4, s₁₇ = 25.63

Sample 18: xbar₁₈ = 244.2, s₁₈ = 29.19

Sample 19: xbar₁₉ = 255.8, s₁₉ = 29.00

Sample 20: xbar₂₀ = 243.9, s₂₀ = 30.17 (click me)

(a) Use Excel to show the distribution of xbars taken from the twenty samples with a histogram. What is it's approximate shape?

The shape is approximately normal. If more xbars were used, it would be normal.

(b) What is the mean and median of the xbars using Excel?

The mean is 259.19 and the median is 250.1. When the means equals the median the data are symmetrical.

(c) What is the standard deviation of the xbars using Excel?

The standard deviation (standard error of the mean) is 7.05.

(d) How are the results related to the Central Limit theorem?

The CLT states that the distribution of xbars taken from a population with mean, m , and standard deviation, s , will be normally distributed with a mean of the xbars equal to m _xbar = m and a standard deviation (standard error), s _xbar= s /Ö (n). In this case, the true population mean is m_xbar = 250 while our sample mean of the xbars was 250.19. The standard deviation (standard error) from the population of xbars is s_xbar= s /Ö(n) = 25/Ö (10) = 7.91 while our sample standard deviation of the xbars was 7.05. Both the sample mean and sample standard deviation would approached those values stated by the CLT as the number of samples increases. The CLT applies to the distribution of xbars from any population as long as the sample size is large (n > 30) and to the distribution of xbars from an normal population at any sample size. The distribution of xbars is not as wide as the original distribution of xs.

(e) Show the normal distribution of the battery lives and associated normal distribution of the average battery lives.

Normal Distribution of Xs ( from the of the population of batteries with normally distributed life times):

Normal Distribution of xbars (average the life times of randomly selected batteries, n = 10, s_xbar = 7.91):

What happens to the size of the standard error of the mean as the sample size increases? Increases Decreases (click one)

Example 2: Conversion of xbars to Z Scores

Show how Z scores are derived from a distribution of xbars. Use three values of xbars: xbar_o = m , xbar₁ = m + s / Ö n and xbar₁ = m - s / Ö n.

(1) Distribution of xbars from Any Distribution of Xs:

(2) Distribution of Z scores:

(click me)

(a) If xbar_o = m,

Z_o = [xbar_o - m] / [s / Ö n]

= [m - m] / [s / Ö n]

= 0 / [s /Ö n ]

= 0

(b) If xbar₁ = m + [s / Ö n],

Z₁ = [xbar₁ - m] / [s / Ö n]

= [ (m +s /Ö n ) - m] / [s / Ö n]

= [s /Ö n ] / [s / Ö n]

= 1

(c) If xbar₂ = m - [s / Ö n],

Z₂ = [xbar₂ - m] / [s / Ö n]

= [ (m - s /Ö n ) - µ] / [s / Ö n]

= [ -s /Ö n ] / [s / Ö n]

= -1

These are only three examples. Any xbar can be standardized (or normalized) to give its own Z score. Just as in the case of converting xs to Z scores, the Z scores associated with xbars count the number of standard deviations (standard errors) that the xbars are way from the m. One uses these Z scores to calculate the probability of a range of xbars.

Given: m = 250, xbar = 275 and SEMean = 7.91 (from Example 1 above). What is the Z score for xbar = 275? 1.00 -1.00 3.16 -3.16 (click one)

Example 3: Probability of x and xbar

Four workers produce hand made widgets. The number of widgets produced is normally distributed with a mean of 24 and a standard deviation of 5.

n = 4

m = 24

s = 5

s_xbar = s / Ö n = 5 / Ö 4 ) = 2.50

(a) What percentage of the time does a worker produce more than 26 widgets per hour? Use Excel to confirm you answer. (click me)

P[x > 26] = P[Z ³ (x - m) / (s )] = P[Z ³ ( 26 - 24 ) / ( 5 )]

= P[Z ³ 0.40 ] = 0.5 - 0.1554 = 0.3446 (confirm in Z table)

Why do we not use the sample size, n = 4, in this problem? Not needed We probably made a mistake (click one)

(b) What percentage of the time is the average rate of output of the 4 workers more than 26 components per hour? Use Excel to confirm you answer. (click me)

P[xbar > 26] = P[Z ³ (Xbar - m) / (s / Ö n)]

= P[Z ³ ( 26 - 24 ) / ( 2.50 )]

= P[Z ³ 0.80 ] = 0.5 - 0.2881 = 0.2119 (confirm in Z table)

What is the new probability in this problem if the sample size increased from n = 4 to n = 16? 0.1056 -0.1056 (click one)

Example 4: Confidence Interval using Z

Assume a normally distributed population of the life time of batteries with a standard deviation 25 minutes when used in an ordinary toy. A random sample of 10 batteries is selected. If the mean from this sample is 255.00, what is a 99% confidence interval for the mean life time of batteries? Use Excel to confirm you answer.

(A) (1) n = 10 (2) xbar = 255 (3) s = 25

(4) s_xbar = s / Ö n = 25 / Ö 10 = 7.91

(B) (1) a = 0.01 (2) a /2 = 0.005 (3) 0.5 - 0.005 = 0.4950

(4) Z_a/2 = Z_0.005 = + 2.575 (5) - Z_a/2 = - Z_0.005 = - 2.575 (click me)

(C) CL = 1 - a = 1 - 0.01 = 0.99 = 99%

(1) 99% CI for m = xbar ± [Za _/2][s / Ö n]

= 255 ± [ 2.575 ][ 25 /Ö 10 ]

= 255 ± [ 2.575 ][ 7.91 ]

= 255 ± 20.37 (E = Maximum Error)

= 234.63 to 275.37

(2) Diagram of CI:

	E		CI = 99%	E
a /2 = 0.005	- 20.37		xbar	+ 20.37		a /2 = 0.005
234.63 (lower)		255			275.37 (upper)

(D) Three Interpretations of CI:

(1) I am 99% confident that the unknown value of m is between 234.63 and 275.37.

(2) If I repeatedly obtained samples of size 10, then 99% of the resulting CIs would contain m and 1% would not.

(3) I am 99% confident that my estimate of m (xbar = 255 ) is within one Maximum Error ( 20.37 ) of the actual value of m.

What would happen to the width of the confidence interval if alpha increased from a = 0.01to a = 0.05? Increase Decrease (click one)

Example 5: Confidence Interval using t

Assume a normally distributed population of the life time of batteries when used in an ordinary toy. A random sample of 10 batteries is selected. If the mean from this sample is 255.0, and a sample standard deviation of 18.16. what is a 99% confidence interval for the mean life time of batteries? Use Excel to confirm you answer.

n = 10

xbar = 255

s = 18.16

Construct a 99% confidence interval for senior faculty at all mid-size universities.

(A) (1) n = 10 (2) xbar = 255 (3) s = 18.16

(4) s_xbar = s / Ö n = 18.16 / Ö 10 = 5.743

(B) (1) a = 0.01 (2) a /2 = 0.005

(3) df = 10 - 1 = 9

(4) t _{a/2,(df

)} = t _{0.005,( 9 )} = + 3.250 (click me)

(5) - t _{a/2,(df )} = - t _{0.005,( 9 )} = - 3.250

(C) CL = 1 - a = 1 - 0.01 = 0.99 = 99%

(1) 99% CI for m = xbar ± [t _a/2,df][ s / Ö n]

= 255 ± [ 3,250 ][ 18.16/ Ö 10]

= 255 ± [ 3.250 ][ 5.743 ]

= 255 ± 18.66 (E = Maximum Error)

= 236.34 to 273.66

(2) Diagram of CI: (using s and [t _a/2,df])

	E		CI = 99%	E
a /2 = 0.005	- 18.66		xbar	+18.66		a /2 = 0.005
236.34 (lower)		255			273.66 (upper)

(D) Three Interpretations of CI:

(1) I am 99% confident that the unknown value of m is between 236.34 and 273.66.

(2) If I repeatedly obtained samples of size 10, then 99% of the resulting CIs would contain m and 1% would not.

(3) I am 99% confident that my estimate of m (xbar = 255 ) is within one Maximum Error ( 18.66 ) of the actual value of m.

Why is t score used in the calculation of this confidence interval instead of a Z score? We are given: (a) Population Sigma and a large sample size (b) sample standard deviation and a small sample size (click one)

Example 6: Sample Size

A doctor would like to measure the effects of new sleeping pills. From past experiments, a measure used to indicate the effectiveness of similar pills has ranged from 8.5 to 12.2 hours. To be 99% confident, how large a sample would be necessary to estimate the mean effectiveness of the new sleeping pills within 30 minutes?

H = 12.2

L = 8.5

a = 0.01

E = 0.5 (click me)

s is not know, use:

s » [H - L] / 4,

» [ 12.2 - 8.5 ] / 4

» [ 3.7 ] / 4

» 0.925

a = 0.01

a/2 = 0.005

0.5 - 0.005 = 0.4950

Z _a/2 = Z _0.005 = 2.575

n = {[Z_a/2][s] / E}²

= {[ 2.575 ][ 0.925 ] / 0.5 }²

= {[ 2.3819 ] / 0.5 }²

= { 4.7638 }²

= 22.69

» 23 (click me)

What would happen to the sample size if the maximum error increased from E = 0.5 to E = 0.75? Increase Decrease (click one)

You should now:

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E-mail Dr. James V. Pinto at BA501@mail.cba.nau.edu
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