EBK INTRODUCTION TO THE PRACTICE OF STA
EBK INTRODUCTION TO THE PRACTICE OF STA
9th Edition
ISBN: 8220103674638
Author: Moore
Publisher: YUZU
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Chapter 7.1, Problem 33E

(a)

To determine

To graph: The histogram.

(a)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

To draw the histogram, follow the below mentioned steps in Minitab:

Step 1: Enter the data in Minitab and enter variable name as “Diameter.”

Step 2: Go to “Graph” and point on “Histogram.” Click on “Simple” and then press OK.

Step 3: In dialog box that appears, select “Diameter” under the field marked as “Graph variable,” and then click on OK.

The histogram obtained is as follows:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 7.1, Problem 33E , additional homework tip  1

Interpretation: The histogram does not seem to represent a bell-shaped curve.

To determine

To graph: The box plot.

Expert Solution
Check Mark

Explanation of Solution

Graph: To draw the box plot, follow the below mentioned steps in Minitab:

Step 1: Enter the data in Minitab and enter variable name as “Diameter.”

Step 2: Go to GraphBoxplotsimple. Then, click on “Ok.”

Step 3: In dialog box that appears, select “Diameter” under the field marked as “Graph variable,” and then click on OK.

The required box plot is as follows:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 7.1, Problem 33E , additional homework tip  2

Interpretation: From the box plot, the data does not appear to be symmetrical.

To determine

To graph: The normal quantile plot.

Expert Solution
Check Mark

Explanation of Solution

Given: The data for the diameters of 40 randomly sampled trees is provided.

Calculation: To draw the normal quantile plot, follow the below mentioned steps in SPSS.

Step 1: Enter the data of diameter in SPSS and enter the variable name “Diameter.”

Step 2: Go to AnalyzeDescriptive statisticsQ-Q plots.

Step 3: Select “Diameter” under the field marked as “Variables.” Finally, click on “Ok.”

The normal quantile plots are as follows:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 7.1, Problem 33E , additional homework tip  3

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 7.1, Problem 33E , additional homework tip  4

To determine

To explain: The distribution.

Expert Solution
Check Mark

Answer to Problem 33E

Solution: The distribution is not normal as it has two peaks, and the normal quantile plot shows that the distribution is deviated from the normal distribution.

Explanation of Solution

The histogram of the distribution shows that it has two peaks. In the box plot, the upper tail is longer than the lower tail, which indicates its deviation from normality, and the normal quantile plots also show that the data is deviated from the normal distribution as most of the values are away from the normal position.

(b)

To determine

Whether t procedures can be used to calculate 95% confidence interval.

(b)

Expert Solution
Check Mark

Answer to Problem 33E

Solution: The t procedures can be used as the sample is large enough to eliminate the effect of non-normal distribution.

Explanation of Solution

As the data is large enough as compared to the population, it will eliminate the effect of non-normality. So, the t procedures can be used.

(c)

Section 1:

To determine

To find: The margin of error.

(c)

Section 1:

Expert Solution
Check Mark

Answer to Problem 33E

Solution: The margin of error is 5.68.

Explanation of Solution

Calculation:

To calculate the margin of error, first, standard deviation and mean of the sample need to be calculated. To calculate standard deviation and mean, follow the below mentioned steps in Minitab:

Step 1: Enter the data in Minitab and enter variable name as “Diameter.”

Step 2: Go to StatBasicstatisticsDisplay Discriptive statistics.

Step 3: In dialog box that appears, select “Diameter” under the field marked as “Variables,” and then click on “Statistics.”

Step 4: In dialog box that appears, select “None” and then “Mean” and “Standard deviation.” Finally, click on “Ok” on both dialog boxes.

From Minitab result, the standard deviation is 17.71 and the mean is 27.29.

The formula for margin of error is as follows:

Marginoferrortα2×sn

where α is the significance level, s is standard deviation of sample, and n is sample size. So, the margin of error is computed as follows:

Marginoferrort0.025×sn=2.03×17.7140=5.68

Section 2:

To determine

To find: The 95% confidence interval for mean diameter.

Section 2:

Expert Solution
Check Mark

Answer to Problem 33E

Solution: The 95% confidence interval is (21.63, 32.95)_.

Explanation of Solution

Calculation:

The confidence interval is an interval for which there are 95% chances that it contains the population parameter (population mean).

To calculate confidence interval, follow the below mentioned steps in Minitab:

Step 1: Enter the provided data into Minitab and enter variable name as “Diameter.”

Step 2: Go to StatBasicstatistics1-sample t.

Step 3: In the dialog box that appears, select “Diameter” under the field marked as “sample in columns.” Click on “option.”

Step 4: In the dialog box that appears, enter “95.0” in confidence level and select “not equal” in “Alternative hypothesis.”

From Minitab results, 95% confidence interval is (21.63, 32.95)_.

Interpretation: The 95% confidence interval is (21.63, 32.95) that means it is 95% sure that the mean diameter for population will lie between 21.63 and 32.95.

(d)

To determine

Whether the result found in parts (a), (b), and (c) would apply to the trees in the same area.

(d)

Expert Solution
Check Mark

Answer to Problem 33E

Solution: The result would not apply to the trees in the same area because the sample size was small and the data was also not normal.

Explanation of Solution

The sample size must be large to conclude about the population but here the sample size was not large enough to conclude about the trees in the same area. Therefore, the result would not be applied to other trees in the same area.

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Chapter 7 Solutions

EBK INTRODUCTION TO THE PRACTICE OF STA

Ch. 7.1 - Prob. 11UYKCh. 7.1 - Prob. 12UYKCh. 7.1 - Prob. 13UYKCh. 7.1 - Prob. 14ECh. 7.1 - Prob. 15ECh. 7.1 - Prob. 16ECh. 7.1 - Prob. 17ECh. 7.1 - Prob. 18ECh. 7.1 - Prob. 19ECh. 7.1 - Prob. 20ECh. 7.1 - Prob. 21ECh. 7.1 - Prob. 22ECh. 7.1 - Prob. 23ECh. 7.1 - Prob. 24ECh. 7.1 - Prob. 25ECh. 7.1 - Prob. 26ECh. 7.1 - Prob. 27ECh. 7.1 - Prob. 28ECh. 7.1 - Prob. 29ECh. 7.1 - Prob. 30ECh. 7.1 - Prob. 31ECh. 7.1 - Prob. 32ECh. 7.1 - Prob. 33ECh. 7.1 - Prob. 34ECh. 7.1 - Prob. 35ECh. 7.1 - Prob. 36ECh. 7.1 - Prob. 37ECh. 7.1 - Prob. 38ECh. 7.1 - Prob. 39ECh. 7.1 - Prob. 40ECh. 7.1 - Prob. 41ECh. 7.1 - Prob. 42ECh. 7.1 - Prob. 43ECh. 7.1 - Prob. 44ECh. 7.1 - Prob. 45ECh. 7.1 - Prob. 46ECh. 7.1 - Prob. 47ECh. 7.2 - Prob. 48UYKCh. 7.2 - Prob. 49UYKCh. 7.2 - Prob. 50UYKCh. 7.2 - Prob. 51UYKCh. 7.2 - Prob. 52UYKCh. 7.2 - Prob. 53UYKCh. 7.2 - Prob. 54UYKCh. 7.2 - Prob. 55ECh. 7.2 - Prob. 56ECh. 7.2 - Prob. 57ECh. 7.2 - Prob. 58ECh. 7.2 - Prob. 59ECh. 7.2 - Prob. 61ECh. 7.2 - Prob. 62ECh. 7.2 - Prob. 63ECh. 7.2 - Prob. 64ECh. 7.2 - Prob. 65ECh. 7.2 - Prob. 66ECh. 7.2 - Prob. 67ECh. 7.2 - Prob. 68ECh. 7.2 - Prob. 69ECh. 7.2 - Prob. 70ECh. 7.2 - Prob. 71ECh. 7.2 - Prob. 72ECh. 7.2 - Prob. 73ECh. 7.2 - Prob. 74ECh. 7.2 - Prob. 75ECh. 7.2 - Prob. 76ECh. 7.2 - Prob. 77ECh. 7.2 - Prob. 78ECh. 7.2 - Prob. 79ECh. 7.2 - Prob. 80ECh. 7.2 - Prob. 81ECh. 7.2 - Prob. 82ECh. 7.2 - Prob. 83ECh. 7.2 - Prob. 84ECh. 7.2 - Prob. 85ECh. 7.2 - Prob. 86ECh. 7.2 - Prob. 87ECh. 7.2 - Prob. 88ECh. 7.2 - Prob. 89ECh. 7.2 - Prob. 90ECh. 7.2 - Prob. 91ECh. 7.2 - Prob. 92ECh. 7.3 - Prob. 93UYKCh. 7.3 - Prob. 94UYKCh. 7.3 - Prob. 95UYKCh. 7.3 - Prob. 96UYKCh. 7.3 - Prob. 97UYKCh. 7.3 - Prob. 98UYKCh. 7.3 - Prob. 99UYKCh. 7.3 - Prob. 100UYKCh. 7.3 - Prob. 101ECh. 7.3 - Prob. 102ECh. 7.3 - Prob. 103ECh. 7.3 - Prob. 104ECh. 7.3 - Prob. 105ECh. 7.3 - Prob. 106ECh. 7.3 - Prob. 107ECh. 7.3 - Prob. 108ECh. 7.3 - Prob. 109ECh. 7.3 - Prob. 110ECh. 7.3 - Prob. 111ECh. 7.3 - Prob. 112ECh. 7.3 - Prob. 113ECh. 7.3 - Prob. 114ECh. 7.3 - Prob. 115ECh. 7.3 - Prob. 116ECh. 7.3 - Prob. 117ECh. 7.3 - Prob. 118ECh. 7 - Prob. 119ECh. 7 - Prob. 120ECh. 7 - Prob. 121ECh. 7 - Prob. 122ECh. 7 - Prob. 123ECh. 7 - Prob. 124ECh. 7 - Prob. 125ECh. 7 - Prob. 126ECh. 7 - Prob. 127ECh. 7 - Prob. 128ECh. 7 - Prob. 129ECh. 7 - Prob. 130ECh. 7 - Prob. 131ECh. 7 - Prob. 132ECh. 7 - Prob. 133ECh. 7 - Prob. 134ECh. 7 - Prob. 135ECh. 7 - Prob. 136ECh. 7 - Prob. 137ECh. 7 - Prob. 138ECh. 7 - Prob. 139ECh. 7 - Prob. 140ECh. 7 - Prob. 141ECh. 7 - Prob. 142ECh. 7 - Prob. 143ECh. 7 - Prob. 144ECh. 7 - Prob. 145ECh. 7 - Prob. 146ECh. 7 - Prob. 147ECh. 7 - Prob. 148ECh. 7 - Prob. 149E
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