The Practice of Statistics for AP - 4th Edition
The Practice of Statistics for AP - 4th Edition
4th Edition
ISBN: 9781429245593
Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.
Publisher: Macmillan Higher Education
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Chapter 7.2, Problem 41E

(a)

To determine

To calculate: The probability that the proportion in an SRS of 100 orders is as small as the proportion in your sample or smaller.

(a)

Expert Solution
Check Mark

Answer to Problem 41E

The required probability is 0.0918.

Explanation of Solution

Given information:

Percent of orders = 90%

Number of orders = 5000

Number of orders for SRS = 100

Number of orders shipped in time = 86

Formula used:

The mean of the sampling distribution of p^ is, μp^=p .

The standard deviation of the sampling distribution of p^ is, σp^=p(1p)n .

Calculation:

Here,

  p=90%=0.90

  x=86

  n=100

Find the mean of sampling distribution of p^ by dividing the number of successes by the sample size that is, p^=xn

Substitute 86 for x and 100 for n in the above formula p^=xn .

  p^=86100=0.86

For the mean of the sampling distribution of p^ , we can use the formula μp^=p .

Substitute 0.90 for p in the formula μp^=p .

  μp^=0.90

For the standard deviation of the sampling distribution, use the formula σp^=p(1p)n .

Substitute 0.90 for p and 100 for n in the above formula and simplify.

  σp^=0.90(10.90)100=0.90×0.1100=0.09100=0.03

Now, we find the z-score by using the formula z=xμσ .

Substitute 0.86 for x , 0.90 for μ and 0.03 for σ in the formula z=xμσ and simplify.

  z=0.860.900.03=0.040.031.33

We can find the probability by using table A as:

  P(p^0.86%)=P(z<1.33)=0.0918

Hence, the required probability is 0.0918.

(b)

To determine

The explanation that why the result of the sample does not refute the 90% claim.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given information:

Percent of orders = 90%

Number of orders = 5000

Number of orders for SRS = 100

Number of orders shipped in time = 86

Formula used:

The mean of the sampling distribution of p^ is, μp^=p .

The standard deviation of the sampling distribution of p^ is, σp^=p(1p)n .

Here,

  p=90%=0.90

  x=86

  n=100

Find the mean of sampling distribution of p^ by dividing the number of successes by the sample size that is, p^=xn

Substitute 86 for x and 100 for n in the above formula p^=xn .

  p^=86100=0.86

For the mean of the sampling distribution of p^ , we can use the formula μp^=p .

Substitute 0.90 for p in the formula μp^=p .

  μp^=0.90

For the standard deviation of the sampling distribution, use the formula σp^=p(1p)n .

Substitute 0.90 for p and 100 for n in the above formula and simplify.

  σp^=0.90(10.90)100=0.90×0.1100=0.09100=0.03

Now, we find the z-score by using the formula z=xμσ .

Substitute 0.86 for x , 0.90 for μ and 0.03 for σ in the formula z=xμσ and simplify.

  z=0.860.900.03=0.040.031.33

We can find the probability by using table A as:

  P(p^0.86%)=P(z<1.33)=0.0918

Thus, the probability is 0.0918.

The above probability is greater than 0.05.

If the true population proportion is 0.90, then a sample can be obtained with sample proportion of 0.86 because the probability is greater than 0.05.

Hence, it is concluded that the result of the sample does not refute the 90% claim.

Chapter 7 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 7.1 - Prob. 3ECh. 7.1 - Prob. 4ECh. 7.1 - Prob. 5ECh. 7.1 - Prob. 6ECh. 7.1 - Prob. 7ECh. 7.1 - Prob. 8ECh. 7.1 - Prob. 9ECh. 7.1 - Prob. 10ECh. 7.1 - Prob. 11ECh. 7.1 - Prob. 12ECh. 7.1 - Prob. 13ECh. 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. 7.2 - Prob. 1.1CYUCh. 7.2 - Prob. 1.2CYUCh. 7.2 - Prob. 1.3CYUCh. 7.2 - Prob. 1.4CYUCh. 7.2 - Prob. 27ECh. 7.2 - Prob. 28ECh. 7.2 - Prob. 29ECh. 7.2 - Prob. 30ECh. 7.2 - Prob. 31ECh. 7.2 - Prob. 32ECh. 7.2 - Prob. 33ECh. 7.2 - Prob. 34ECh. 7.2 - Prob. 35ECh. 7.2 - Prob. 36ECh. 7.2 - Prob. 37ECh. 7.2 - Prob. 38ECh. 7.2 - Prob. 39ECh. 7.2 - Prob. 40ECh. 7.2 - Prob. 41ECh. 7.2 - Prob. 42ECh. 7.2 - Prob. 43ECh. 7.2 - Prob. 44ECh. 7.2 - Prob. 45ECh. 7.2 - Prob. 46ECh. 7.2 - Prob. 47ECh. 7.2 - Prob. 48ECh. 7.3 - Prob. 1.1CYUCh. 7.3 - Prob. 1.2CYUCh. 7.3 - Prob. 1.3CYUCh. 7.3 - Prob. 1.4CYUCh. 7.3 - Prob. 49ECh. 7.3 - Prob. 50ECh. 7.3 - Prob. 51ECh. 7.3 - Prob. 52ECh. 7.3 - Prob. 53ECh. 7.3 - Prob. 54ECh. 7.3 - Prob. 55ECh. 7.3 - Prob. 56ECh. 7.3 - Prob. 57ECh. 7.3 - Prob. 58ECh. 7.3 - Prob. 59ECh. 7.3 - Prob. 60ECh. 7.3 - Prob. 61ECh. 7.3 - Prob. 62ECh. 7.3 - Prob. 63ECh. 7.3 - Prob. 64ECh. 7.3 - Prob. 65ECh. 7.3 - Prob. 66ECh. 7.3 - Prob. 67ECh. 7.3 - Prob. 68ECh. 7.3 - Prob. 69ECh. 7.3 - Prob. 70ECh. 7.3 - Prob. 71ECh. 7.3 - Prob. 72ECh. 7 - Prob. 1CRECh. 7 - Prob. 2CRECh. 7 - Prob. 3CRECh. 7 - Prob. 4CRECh. 7 - Prob. 5CRECh. 7 - Prob. 6CRECh. 7 - Prob. 7CRECh. 7 - Prob. 1PTCh. 7 - Prob. 2PTCh. 7 - Prob. 3PTCh. 7 - Prob. 4PTCh. 7 - Prob. 5PTCh. 7 - Prob. 6PTCh. 7 - Prob. 7PTCh. 7 - Prob. 8PTCh. 7 - Prob. 9PTCh. 7 - Prob. 10PTCh. 7 - Prob. 11PTCh. 7 - Prob. 12PTCh. 7 - Prob. 13PTCh. 7 - Prob. 1PT2Ch. 7 - Prob. 2PT2Ch. 7 - Prob. 3PT2Ch. 7 - Prob. 4PT2Ch. 7 - Prob. 5PT2Ch. 7 - Prob. 6PT2Ch. 7 - Prob. 7PT2Ch. 7 - Prob. 8PT2Ch. 7 - Prob. 9PT2Ch. 7 - Prob. 10PT2Ch. 7 - Prob. 11PT2Ch. 7 - Prob. 12PT2Ch. 7 - Prob. 13PT2Ch. 7 - Prob. 14PT2Ch. 7 - Prob. 15PT2Ch. 7 - Prob. 16PT2Ch. 7 - Prob. 17PT2Ch. 7 - Prob. 18PT2Ch. 7 - Prob. 19PT2Ch. 7 - Prob. 20PT2Ch. 7 - Prob. 21PT2Ch. 7 - Prob. 22PT2Ch. 7 - Prob. 23PT2Ch. 7 - Prob. 24PT2Ch. 7 - Prob. 25PT2
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