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 5, Problem 95E

(a)

Section 1:

To determine

To find: The distribution of the proportion of working females.

(a)

Section 1:

Expert Solution
Check Mark

Answer to Problem 95E

Solution: The proportion of females who were engaged in work follows normal distribution with mean μp^F=0.82 and standard deviation σp^F=0.0192.

Explanation of Solution

Calculation: The sampling distribution of the sample proportion of success p^ is distributed normally with mean μp^=p and standard deviation σp^=p(1p)n, where p is the success proportion in the population and n is the size of sample.

Therefore, for the given problem,

μp^F=pF=0.82

And,

σp^F=pF(1pF)nF=0.82×(10.82)400=0.0192

Section 2:

To determine

To find: The distribution of the proportion of working males.

Section 2:

Expert Solution
Check Mark

Answer to Problem 95E

Solution: The proportion of males who were engaged in work follows normal distribution with average μp^M=0.88 and standard deviation σp^M=0.01625.

Explanation of Solution

Calculation: The sampling distribution of the sample proportion of success p^ is distributed normally with mean μp^=p and standard deviation σp^=p(1p)n. Therefore, for the given problem,

μp^M=pM=0.88

And,

σp^M=pM(1pM)nM=0.88×(10.88)400=0.01625

(b)

To determine

To find: The sampling distribution of the difference of sample proportions.

(b)

Expert Solution
Check Mark

Answer to Problem 95E

Solution: The sampling distribution of (p^Mp^F) is distributed normally with mean μp^Mp^F=0.06 and standard deviation σp^Mp^F=0.02516.

Explanation of Solution

Calculation: The proportion of females who were engaged in work follows normal distribution with mean μp^F=0.82 and standard deviation σp^F=0.01921.

The proportion of males who were engaged in work follows normal distribution with mean μp^M=0.88 and standard deviation σp^M=0.01625.

The sampling distribution of p^Mp^F is distributed normally with mean μp^Mp^F=p^Mp^F and standard deviation σp^Mp^F=σpM2+σpF2. Therefore,

μp^Mp^F=p^Mp^F=0.880.82=0.06

And,

σp^Mp^F=σpM2+σpF2=0.019212+0.016252=0.02516

(c)

To determine

To find: The probability that a higher proportion of females worked than males.

(c)

Expert Solution
Check Mark

Answer to Problem 95E

Solution: The required probability is P(p^F>p^M)=0.0087_.

Explanation of Solution

Calculation: The following probability needs to be calculated for the given problem:

P(p^M<p^F)=P(p^Mp^F<0)

The sampling distribution of p^Mp^F is distributed normally with mean μp^Mp^F=0.06 and standard deviation σp^Mp^F=0.02516.

The probability that p^Mp^F is smaller than 0, which is the area to the left of the point 0 under the normal curve. To find the area under the normal curve, first convert the value of sample mean p^Mp^F=0 to a Z-score. The area left to the particular Z-score can be obtained using Table A provided in the book.

The Z-score for the value p^Mp^F=0 is calculated as:

Z=(p^Mp^F)μp^Mp^Fσp^Mp^F=00.060.02516=2.38

The area below the standard normal curve to the left of Z=2.38 is obtained as 0.0087 from the standard normal table.

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Chapter 5 Solutions

EBK INTRODUCTION TO THE PRACTICE OF STA

Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.2 - Prob. 17UYKCh. 5.2 - Prob. 18UYKCh. 5.2 - Prob. 19UYKCh. 5.2 - Prob. 20UYKCh. 5.2 - Prob. 21UYKCh. 5.2 - Prob. 22UYKCh. 5.2 - Prob. 23UYKCh. 5.2 - Prob. 24UYKCh. 5.2 - Prob. 25ECh. 5.2 - Prob. 26ECh. 5.2 - Prob. 27ECh. 5.2 - Prob. 28ECh. 5.2 - Prob. 29ECh. 5.2 - Prob. 30ECh. 5.2 - Prob. 31ECh. 5.2 - Prob. 32ECh. 5.2 - Prob. 33ECh. 5.2 - Prob. 34ECh. 5.2 - Prob. 35ECh. 5.2 - Prob. 36ECh. 5.2 - Prob. 37ECh. 5.2 - Prob. 38ECh. 5.2 - Prob. 39ECh. 5.2 - Prob. 40ECh. 5.2 - Prob. 41ECh. 5.2 - Prob. 42ECh. 5.3 - Prob. 43UYKCh. 5.3 - Prob. 44UYKCh. 5.3 - Prob. 45UYKCh. 5.3 - Prob. 46UYKCh. 5.3 - Prob. 47UYKCh. 5.3 - Prob. 48UYKCh. 5.3 - Prob. 49UYKCh. 5.3 - Prob. 50UYKCh. 5.3 - Prob. 51UYKCh. 5.3 - Prob. 52UYKCh. 5.3 - Prob. 53UYKCh. 5.3 - Prob. 54UYKCh. 5.3 - Prob. 55UYKCh. 5.3 - Prob. 56UYKCh. 5.3 - Prob. 57ECh. 5.3 - Prob. 58ECh. 5.3 - Prob. 59ECh. 5.3 - Prob. 60ECh. 5.3 - Prob. 61ECh. 5.3 - Prob. 62ECh. 5.3 - Prob. 63ECh. 5.3 - Prob. 64ECh. 5.3 - Prob. 65ECh. 5.3 - Prob. 66ECh. 5.3 - Prob. 67ECh. 5.3 - Prob. 68ECh. 5.3 - Prob. 69ECh. 5.3 - Prob. 70ECh. 5.3 - Prob. 71ECh. 5.3 - Prob. 72ECh. 5.3 - Prob. 73ECh. 5.3 - Prob. 74ECh. 5.3 - Prob. 75ECh. 5.3 - Prob. 76ECh. 5.3 - Prob. 77ECh. 5.3 - Prob. 78ECh. 5.3 - Prob. 79ECh. 5.3 - Prob. 80ECh. 5.3 - Prob. 81ECh. 5.3 - Prob. 82ECh. 5.3 - Prob. 83ECh. 5 - Prob. 84ECh. 5 - Prob. 85ECh. 5 - Prob. 86ECh. 5 - Prob. 87ECh. 5 - Prob. 88ECh. 5 - Prob. 89ECh. 5 - Prob. 90ECh. 5 - Prob. 91ECh. 5 - Prob. 92ECh. 5 - Prob. 93ECh. 5 - Prob. 94ECh. 5 - Prob. 95ECh. 5 - Prob. 96ECh. 5 - Prob. 97ECh. 5 - Prob. 98ECh. 5 - Prob. 99ECh. 5 - Prob. 100ECh. 5 - Prob. 101ECh. 5 - Prob. 102E
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