Essential Statistics
Essential Statistics
2nd Edition
ISBN: 9781259570643
Author: Navidi
Publisher: MCG
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Chapter 6.2, Problem 19E

a.

To determine

Find the population mean μ and standard deviation σ.

a.

Expert Solution
Check Mark

Answer to Problem 19E

The population mean μ and standard deviation σ are 75.4 and 5.1225.

Explanation of Solution

Calculation:

The given data shows that the summer temperature for five days in July.

The formula to find the population mean is,

 μ=Sum of temperaturePopulation size=69+75+79+83+715=3775=75.4

Thus, the population mean μ is 75.4.

The formula to find the population standard deviation is,

σ=(xμ)2N

Here, x represents the population values,µ represents the population mean and N represents the population size.

Substitute µ as75.4 and N as5.

That is,

σ=(6975.4)2+(7575.4)2+(7975.4)2+(8375.4)2+(7175.4)25=(6.4)2+(0.4)2+(3.6)2+(7.6)2+(4.4)25=40.96+0.16+12.96+57.76+19.365=131.25

=26.24=5.1225

Thus, the standard deviation σ is 5.1225.

b.

To determine

List all samples of size 2 drawn with replacement.

b.

Expert Solution
Check Mark

Explanation of Solution

Calculation:

The two values are randomly selected with replacement from the population {69, 75, 79, 83, 71}.

Thedifferent possible samples of size 2 are {(69, 69), (69,75), (69, 79), (69, 83), (69, 71), (75, 69), (75, 75), (75, 79), (75, 83), (75, 71), (79, 69), (79, 75), (79, 79), (79, 83), (79, 71), (83, 69), (83, 75), (83, 79), (83, 83), (83, 71), (71, 69), (71, 75), (71, 79), (71, 83), (71, 71)}.

Thus, there are 25 different samples of size 2.

c.

To determine

Compute the sample mean x¯ for each of the 25 samples of size 2. Also, compute the mean μx¯ and standard deviation σx¯ of the samples means.

c.

Expert Solution
Check Mark

Answer to Problem 19E

The sample mean x¯ for each of the 25 samples of size 2 are,

SampleSample mean (x¯)
69, 6969
69, 7572
69, 7974
69, 8376
69, 7170
75, 6972
75, 7575
75, 7977
75, 8379
75, 7173
79, 6974
79, 7577
79, 7979
79, 8381
79, 7175
83, 6976
83, 7579
83, 7981
83, 8383
83, 7177
71, 6970
71, 7573
71, 7975
71, 8377
71, 7171
Total1,885

The mean μx¯ and standard deviation σx¯ of the samples means are 75.4 and 3.62215.

Explanation of Solution

Calculation:

The formula for finding sample mean is x¯ is,

x¯=Sum of samplesSample size

The sample mean for the sample (69, 69):

x¯=69+692=1382=69

The formula to find the sample variance is,

Sample standeard deviation=(xx¯)2n1

Where, x represents the sample values, x¯ represents the sample mean and n represents the sample size.

The sample variance for the sample (69, 69):

Substitute x¯ as 69 and n as 2 in sample standard deviation

That is,

Sample variance=(6969)2+(6969)221=(0)2+(0)21=0+01=0

Similarly, the sample mean and sample standard deviation for the remaining samples of size 2 is obtained as given in below Table.

SampleSample mean (x¯)Sample variance
69, 69690
69, 757218
69, 797450
69, 837698
69, 71702
75, 697218
75, 75750
75, 79778
75, 837932
75, 71738
79, 697450
79, 75778
79, 79790
79, 83818
79, 717532
83, 697698
83, 757932
83, 79818
83, 83830
83, 717772
71, 69702
71, 75738
71, 797532
71, 837772
71, 71710
Total1,885656

The mean μx¯ of the sample means is obtained as follows:

μx¯=Sum of sample meansTotal number of samples=1,88525=75.4

Thus, the mean μx¯ of the sample means is 75.4.

The standard deviation σx¯ of the sample means is obtained as follows:

σx¯=σ22n=65650=3.62215

Thus, the standard deviation σx¯ of the sample means is 3.62215.

d.

To determine

Verify that μx¯=μ and σx¯=σn.

d.

Expert Solution
Check Mark

Explanation of Solution

Calculation:

The two values are randomly selected with replacement. Therefore, n is 2.

From part (a), the population mean μ and standard deviation σ are 75.4 and 5.1225.

Substitute 75.4 for μ in μx¯=μ

μx¯=75.4

Substitute 5.1225 for σ and 2 for n in σx¯=σn

σx¯=5.12252=5.12251.4142135=3.62215

From part (c), the mean μx¯ and standard deviation σx¯ of the samples means are 75.4 and 3.62215.

Thus, the mean and standard deviation both are equal.

Hence μx¯=μ and σx¯=σn.

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Chapter 6 Solutions

Essential Statistics

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