PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Chapter 9.2, Problem 46E

(a)

To determine

To Explain: the type-I error and Type-II error in this setting and give a possible consequence of each.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Claim is Proportion is less than 10%

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis statement is that the population proportion is equal to the value given in the claim. If the null hypothesis is the claim, then the alternative hypothesis statement is the opposite of the null hypothesis.

  H0:p=10%=0.10H1:p<0.10

Type I error is the rejecting the null hypothesis H0 , once the null hypothesis H0 is true.

There is enough convincing proof that the percentage of patients who experience nausea is less than 10%, whereas the percentage of patients who experience nausea is actually 10%.

A possible consequence is that the doctors underestimate the percentage of patients who experience nausea and therefore the side effects are worse then they expect.

Type II error is fails to reject the null hypothesis H0 , once the null hypothesis H0 is false.

There is sufficient convincing evidence that the percentage of patients who experience nausea is less than 10%, while the percentage of patients who experience nausea is actually 10%.

A possible consequence is that the doctors overestimate the percentage of patients who experience nausea and thus the side effects are better then they expect, which could lead to an effective drug not being put on the market (due to the bad side effects).

(b)

To determine

To Explain: these data provide convincing proof for the drug manufacturer’s claim.

(b)

Expert Solution
Check Mark

Answer to Problem 46E

There is no enough convincing evidence that the drug manufacturer’s claim is false.

Explanation of Solution

Given:

  α=0.05n=300x=25

Claim is less than 10%

Formula used:

  p^=xnz=p^p0p0(1p0)n

Calculation:

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis statement is that the population proportion is equal to the value given in the claim. If the null hypothesis is the claim, then the alternative hypothesis statement is the opposite of the null hypothesis.

  H0:p=10%=0.10H1:p<0.10

Normal: Satisfied, because

  np0=300(0.10)=30and n(1p0)=300(10.10)=300(0.90)=270

are both at least 10.

Since all conditions are satisfied, it is appropriate to use a hypothesis test for the population proportion p .

The sample proportion is

  p^=xn=25300=112=0.0833

The test-statistic is

  z=p^p0p0(1p0)n=0.08330.100.10(10.10)300=0.96

The P-value is the probability of getting the value of the test statistic, or a value more extreme, when the null hypothesis is true. find the P-value using the normal probability table

  P=P(Z<0.96)=0.1685

If the P-value is lesser than the significance level α , then reject the null hypothesis:

  P>0.05Fail to rejectH0 

There is no enough convincing evidence that the drug manufacturer’s claim is false.

Chapter 9 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 9.1 - Prob. 11ECh. 9.1 - Prob. 12ECh. 9.1 - Prob. 13ECh. 9.1 - Prob. 14ECh. 9.1 - Prob. 15ECh. 9.1 - Prob. 16ECh. 9.1 - Prob. 17ECh. 9.1 - Prob. 18ECh. 9.1 - Prob. 19ECh. 9.1 - Prob. 20ECh. 9.1 - Prob. 21ECh. 9.1 - Prob. 22ECh. 9.1 - Prob. 23ECh. 9.1 - Prob. 24ECh. 9.1 - Prob. 25ECh. 9.1 - Prob. 26ECh. 9.1 - Prob. 27ECh. 9.1 - Prob. 28ECh. 9.1 - Prob. 29ECh. 9.1 - Prob. 30ECh. 9.1 - Prob. 31ECh. 9.1 - Prob. 32ECh. 9.1 - Prob. 33ECh. 9.1 - Prob. 34ECh. 9.2 - Prob. 35ECh. 9.2 - Prob. 36ECh. 9.2 - Prob. 37ECh. 9.2 - Prob. 38ECh. 9.2 - Prob. 39ECh. 9.2 - Prob. 40ECh. 9.2 - Prob. 41ECh. 9.2 - Prob. 42ECh. 9.2 - Prob. 43ECh. 9.2 - Prob. 44ECh. 9.2 - Prob. 45ECh. 9.2 - Prob. 46ECh. 9.2 - Prob. 47ECh. 9.2 - Prob. 48ECh. 9.2 - Prob. 49ECh. 9.2 - Prob. 50ECh. 9.2 - Prob. 51ECh. 9.2 - Prob. 52ECh. 9.2 - Prob. 53ECh. 9.2 - Prob. 54ECh. 9.2 - Prob. 55ECh. 9.2 - Prob. 56ECh. 9.2 - Prob. 57ECh. 9.2 - Prob. 58ECh. 9.2 - Prob. 59ECh. 9.2 - Prob. 60ECh. 9.2 - Prob. 61ECh. 9.2 - Prob. 62ECh. 9.2 - Prob. 63ECh. 9.2 - Prob. 64ECh. 9.3 - Prob. 65ECh. 9.3 - Prob. 66ECh. 9.3 - Prob. 67ECh. 9.3 - Prob. 68ECh. 9.3 - Prob. 69ECh. 9.3 - Prob. 70ECh. 9.3 - Prob. 71ECh. 9.3 - Prob. 72ECh. 9.3 - Prob. 73ECh. 9.3 - Prob. 74ECh. 9.3 - Prob. 75ECh. 9.3 - Prob. 76ECh. 9.3 - Prob. 77ECh. 9.3 - Prob. 78ECh. 9.3 - Prob. 79ECh. 9.3 - Prob. 80ECh. 9.3 - Prob. 81ECh. 9.3 - Prob. 82ECh. 9.3 - Prob. 83ECh. 9.3 - Prob. 84ECh. 9.3 - Prob. 85ECh. 9.3 - Prob. 86ECh. 9.3 - Prob. 87ECh. 9.3 - Prob. 88ECh. 9.3 - Prob. 89ECh. 9.3 - Prob. 90ECh. 9.3 - Prob. 91ECh. 9.3 - Prob. 92ECh. 9.3 - Prob. 93ECh. 9.3 - Prob. 94ECh. 9.3 - Prob. 95ECh. 9.3 - Prob. 96ECh. 9.3 - Prob. 97ECh. 9.3 - Prob. 98ECh. 9.3 - Prob. 99ECh. 9.3 - Prob. 100ECh. 9.3 - Prob. 101ECh. 9.3 - Prob. 102ECh. 9.3 - Prob. 103ECh. 9.3 - Prob. 104ECh. 9.3 - Prob. 105ECh. 9.3 - Prob. 106ECh. 9.3 - Prob. 107ECh. 9.3 - Prob. 108ECh. 9.3 - Prob. 109ECh. 9.3 - Prob. 110ECh. 9 - Prob. R9.1RECh. 9 - Prob. R9.2RECh. 9 - Prob. R9.3RECh. 9 - Prob. R9.4RECh. 9 - Prob. R9.5RECh. 9 - Prob. R9.6RECh. 9 - Prob. R9.7RECh. 9 - Prob. T9.1SPTCh. 9 - Prob. T9.2SPTCh. 9 - Prob. T9.3SPTCh. 9 - Prob. T9.4SPTCh. 9 - Prob. T9.5SPTCh. 9 - Prob. T9.6SPTCh. 9 - Prob. T9.7SPTCh. 9 - Prob. T9.8SPTCh. 9 - Prob. T9.9SPTCh. 9 - Prob. T9.10SPTCh. 9 - Prob. T9.11SPTCh. 9 - Prob. T9.12SPTCh. 9 - Prob. T9.13SPT
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