Essential Statistics
Essential Statistics
2nd Edition
ISBN: 9781259570643
Author: Navidi
Publisher: MCG
Question
Chapter 9.2, Problem 1CYU

a.

To determine

State the null and alternate hypotheses.

a.

Expert Solution
Check Mark

Answer to Problem 1CYU

The hypotheses are given below:

Null hypothesis:

H0:p1=p2

That is, there is no significant difference between the proportion of patients experienced relief from pain under drug-1 and proportion of patients experienced relief from pain under drug-2.

Alternate hypothesis:

H1:p1>p2

That is, the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2.

Explanation of Solution

It is given that among a sample of 100 patients under drug-1, 76 patients experienced relief from pain and among a sample of 200 patients under drug-2, 128 patients experienced relief from pain. The investigator wants to check whether the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2. The level of significance is α=0.05.

Hypothesis:

Hypothesis is an assumption about the parameter of the population, and the assumption may or may not be true.

Let p1 be the population proportion of patients experienced relief from pain under drug-1 and p2 be the population proportion of patients experienced relief from pain under drug-2.

Claim:

Here, the claim is whether the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2.

The hypotheses are given below:

Null hypothesis:

Null hypothesis is a statement which is tested for statistical significance in the test. The decision criterion indicates whether the null hypothesis will be rejected or not in the favor of alternate hypothesis.

H0:p1=p2

That is, there is no significant difference between the proportion of patients experienced relief from pain under drug-1 and proportion of patients experienced relief from pain under drug-2.

Alternate hypothesis:

Alternate hypothesis is contradictory statement of the null hypothesis

H1:p1>p2

That is, the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2.

b.

To determine

Find the proportion of patients experienced relief from pain under drug-1.

Find the proportion of patients experienced relief from pain under drug-2.

b.

Expert Solution
Check Mark

Answer to Problem 1CYU

The proportion of patients experienced relief from pain under drug-1 is p^1=0.76_.

The proportion of patients experienced relief from pain under drug-2 is p^2=0.64_

Explanation of Solution

Calculation:

Proportion of patients experienced relief from pain under drug-1:

The total number of patients under drug-1 is n1=100 and the number of patients experienced relief from pain under drug-1 is x1=76.

The proportion of patients experienced relief from pain under drug-1 is obtained as follows:

p^1=x1n1=76100=0.76

Thus, the proportion of patients experienced relief from pain under drug-1 is 0.76.

Proportion of patients experienced relief from pain under drug-2:

The total number of patients under drug-2 is n2=200 and the number of patients experienced relief from pain under drug-2 is x2=128.

The proportion of patients experienced relief from pain under drug-2 is obtained as follows:

p^2=x2n2=128200=0.64

Thus, the proportion of patients experienced relief from pain under drug-2 is 0.64.

c.

To determine

Find the value of t-test statistic.

c.

Expert Solution
Check Mark

Answer to Problem 1CYU

The value of test statistic is 2.100436.

Explanation of Solution

Calculation:

Estimate of pooled proportion:

The estimate of pooled proportion is obtained as follows:

p^=x1+x2n1+n2=76+128100+200=204300=0.68

Thus, the estimate of pooled proportion is 0.68.

From part (b), the sample proportions are p^1=0.76 and p^2=0.64.

Test statistic:

The test statistic for testing the difference between two proportions is,

z=(p^1p^2)(p1p2)p^(1p^)(1n1+1n2)

Under the null hypothesis, (p1p2)=0.

The test statistic is obtained as follows,

z=(p^1p^2)0p^(1p^)(1n1+1n2)=0.760.640.68(10.68)(1100+1200)=0.120.057131=2.100436

Thus, the test statistic is 2.100436.

d.

To determine

Find the P-value for the test statistic.

d.

Expert Solution
Check Mark

Answer to Problem 1CYU

The P-value for the test statistic is 0.01785.

Explanation of Solution

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

  • Choose Graph > Probability Distribution Plot.
  • Choose View Probability > OK.
  • From Distribution, choose ‘Normal’ distribution.
  • Click the Shaded Area tab.
  • Choose X value and Right Tail for the region of the curve to shade.
  • In X-value enter 2.100436.
  • Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.2, Problem 1CYU

From the MINITAB output, the P-value is 0.01785.

Thus, the P-value is 0.01785.

e.

To determine

Interpret the P-value at the level of significance α=0.05.

e.

Expert Solution
Check Mark

Answer to Problem 1CYU

There is enough evidence to reject the null hypothesis H0.

Explanation of Solution

From part (d), the P-value is 0.01785.

Decision rule based on P-value:

If Pvalueα, then reject the null hypothesis H0.

If Pvalue>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.05.

Conclusion based on P-value approach:

The P-value is 0.01785 and α value is 0.05.

Here, P-value is less than the α value.

That is, 0.01785(=Pvalue)<0.05(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

f.

To determine

State the conclusion.

f.

Expert Solution
Check Mark

Answer to Problem 1CYU

There is enough evidence to conclude that the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2.

Explanation of Solution

From part (e), it is known that the null hypothesis is rejected.

Thus, there is not enough evidence to conclude that the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2.

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Chapter 9 Solutions

Essential Statistics

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 - 27. Does this diet help? A group of 78 people...Ch. 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. 9.1 - Prob. 35ECh. 9.1 - Prob. 36ECh. 9.1 - Prob. 37ECh. 9.1 - 38. Interpret calculator display: The following...Ch. 9.1 - Prob. 39ECh. 9.1 - Prob. 40ECh. 9.1 - Prob. 41ECh. 9.2 - Prob. 1CYUCh. 9.2 - Prob. 2CYUCh. 9.2 - Prob. 3CYUCh. 9.2 - Prob. 4CYUCh. 9.2 - Prob. 5ECh. 9.2 - Prob. 6ECh. 9.2 - Prob. 7ECh. 9.2 - Prob. 8ECh. 9.2 - Prob. 9ECh. 9.2 - Prob. 10ECh. 9.2 - Prob. 11ECh. 9.2 - Prob. 12ECh. 9.2 - Prob. 13ECh. 9.2 - Prob. 14ECh. 9.2 - Prob. 15ECh. 9.2 - Prob. 16ECh. 9.2 - Prob. 17ECh. 9.2 - Prob. 18ECh. 9.2 - Prob. 19ECh. 9.2 - Prob. 20ECh. 9.2 - Prob. 21ECh. 9.2 - Prob. 22ECh. 9.2 - Prob. 23ECh. 9.2 - Prob. 24ECh. 9.2 - Prob. 25ECh. 9.2 - Prob. 26ECh. 9.2 - Prob. 27ECh. 9.2 - Prob. 28ECh. 9.2 - Prob. 29ECh. 9.2 - Prob. 30ECh. 9.2 - Prob. 31ECh. 9.2 - Prob. 32ECh. 9.2 - Prob. 33ECh. 9.2 - Prob. 34ECh. 9.2 - Prob. 35ECh. 9.2 - Prob. 36ECh. 9.2 - Prob. 37ECh. 9.3 - Prob. 1CYUCh. 9.3 - Prob. 2CYUCh. 9.3 - Prob. 3CYUCh. 9.3 - Prob. 4CYUCh. 9.3 - Prob. 5ECh. 9.3 - Prob. 6ECh. 9.3 - Prob. 7ECh. 9.3 - Prob. 8ECh. 9.3 - Prob. 9ECh. 9.3 - Prob. 10ECh. 9.3 - Prob. 11ECh. 9.3 - Prob. 12ECh. 9.3 - Prob. 13ECh. 9.3 - Prob. 14ECh. 9.3 - Prob. 15ECh. 9.3 - SAT coaching: A sample of 32 students took a class...Ch. 9.3 - Prob. 17ECh. 9.3 - Prob. 18ECh. 9.3 - Prob. 19ECh. 9.3 - Tires and fuel economy: A tire manufacturer is...Ch. 9.3 - Prob. 21ECh. 9.3 - Prob. 22ECh. 9.3 - Prob. 23ECh. 9.3 - Prob. 24ECh. 9.3 - Prob. 25ECh. 9.3 - Prob. 26ECh. 9.3 - Prob. 27ECh. 9.3 - Prob. 28ECh. 9.3 - Prob. 29ECh. 9.3 - Prob. 30ECh. 9.3 - Advantage of matched pairs: Refer to Exercise...Ch. 9.3 - Prob. 32ECh. 9 - A sample of 15 weight litters is tested to see how...Ch. 9 - Prob. 2CQCh. 9 - Prob. 3CQCh. 9 - Prob. 4CQCh. 9 - Prob. 5CQCh. 9 - Prob. 6CQCh. 9 - Prob. 7CQCh. 9 - Prob. 8CQCh. 9 - Prob. 9CQCh. 9 - Prob. 10CQCh. 9 - Prob. 11CQCh. 9 - In a survey of 300 randomly selected female and...Ch. 9 - Prob. 13CQCh. 9 - Prob. 14CQCh. 9 - Prob. 15CQCh. 9 - Prob. 1RECh. 9 - Prob. 2RECh. 9 - Prob. 3RECh. 9 - Prob. 4RECh. 9 - Prob. 5RECh. 9 - Prob. 6RECh. 9 - Prob. 7RECh. 9 - Prob. 8RECh. 9 - Prob. 9RECh. 9 - Polling results: A simple random sample of 400...Ch. 9 - Treating bean plants: In a study to measure the...Ch. 9 - Prob. 12RECh. 9 - Prob. 13RECh. 9 - Prob. 14RECh. 9 - Prob. 15RECh. 9 - Prob. 1WAICh. 9 - Prob. 2WAICh. 9 - Describe the differences between performing a...Ch. 9 - Prob. 4WAICh. 9 - In what ways is the procedure for constructing a...Ch. 9 - Prob. 6WAICh. 9 - Prob. 7WAICh. 9 - Prob. 1CSCh. 9 - Prob. 2CSCh. 9 - Prob. 3CSCh. 9 - Prob. 4CSCh. 9 - Prob. 5CSCh. 9 - Prob. 6CS
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