Essential Statistics
Essential Statistics
2nd Edition
ISBN: 9781259570643
Author: Navidi
Publisher: MCG
Question
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Chapter 9.2, Problem 37E

a.

To determine

State the null and alternate hypotheses.

a.

Expert Solution
Check Mark

Answer to Problem 37E

The hypotheses are given below:

Null hypothesis:

H0:(p1p2)0.05.

Alternate hypothesis:

H1:(p1p2)>0.05.

Explanation of Solution

It is given that among a sample of 500 chips produced by less expensive machine, 70 chips were defective and among a sample of 400 chips produced by more expensive machine, 20 chips were defective. The manufacturer wants to buy more expensive machine if the proportion of defectives is more than 5% less than on the less expensive machine.

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 defectives produced by the less expensive machine and p2 be the population proportion of defectives produced by the more expensive machine.

Claim:

Here, the claim is that the manufacturer wants to buy more expensive machine if the proportion of defectives is more than 5% less than on the less expensive machine.

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:(p1p2)0.05

That is, the proportion of defectives produced by less expensive machine is less than or equal to 5% more than the proportion of defectives produced by more expensive machine.

Alternate hypothesis:

Alternate hypothesis is contradictory statement of the null hypothesis

H1:(p1p2)>0.05

That is, the proportion of defectives produced by less expensive machine is more than 5% more than the proportion of defectives produced by more expensive machine.

b.

To determine

Find the value of test statistic.

b.

Expert Solution
Check Mark

Answer to Problem 37E

The value of test statistic is 4.4776.

Explanation of Solution

Calculation:

The hypotheses are given below:

Null hypothesis:

H0:(p1p2)0.05

That is, the proportion of defectives produced by less expensive machine is less than or equal to 5% more than the proportion of defectives produced by more expensive machine.

Alternate hypothesis:

H1:(p1p2)>0.05

That is, the proportion of defectives produced by less expensive machine is more than 5% more than the proportion of defectives produced by more expensive machine.

Point estimates:

The total number of chips produced by less expensive machine is n1=500 and the number of defectives by the less expensive machine is x1=70.

The point estimate for the proportion of defectives produced by the less expensive machine is obtained as follows:

p^1=x1n1=70500=0.14

Thus, the point estimate for the proportion of defectives produced by the less expensive machine is 0.14.

The total number of chips produced by more expensive machine is n2=400 and the number of defectives by the more expensive machine is x2=20.

The point estimate for the proportion of defectives produced by the more expensive machine is obtained as follows:

p^2=x2n2=20400=0.05

Thus, the point estimate for the proportion of defectives produced by the more expensive machine is 0.05.

Estimate of pooled proportion:

The estimate of pooled proportion is obtained as follows:

p^=x1+x2n1+n2=70+20500+400=90900=0.1

Thus, the estimate of pooled proportion is 0.1.

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)=0.140.050.1(10.1)(1500+1400)=0.090.0201=4.4776

Thus, the test statistic is 4.4776.

c.

To determine

Check whether the null hypothesis is rejected at α=0.05.

c.

Expert Solution
Check Mark

Answer to Problem 37E

The null hypothesis is rejected at α=0.05.

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.
  • Under Distribution, choose ‘Normal’ distribution.
  • Enter 0 as mean and 1 as standard deviation.
  • Click the Shaded Area tab.
  • Under Define shaded area by choose X value and Right Tail.
  • In X-value enter 4.4776.
  • Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.2, Problem 37E

From the MINITAB output, the P-value is 3.7743×106.

Thus, the P-value is 3.7743×106.

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 3.7743×106 and α value is 0.05.

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

That is, 3.7743×106(=Pvalue)<0.05(=α).

By the rejection rule, reject the null hypothesis.

Thus, the null hypothesis is rejected.

d.

To determine

Find the machine that has to be bought by manufacturer.

d.

Expert Solution
Check Mark

Answer to Problem 37E

The manufacturer has to by the more expensive machine.

Explanation of Solution

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

That is, the proportion of defectives produced by less expensive machine is more than 5% more than the proportion of defectives produced by more expensive machine.

The claim was that, the manufacturer wants to buy more expensive machine if the proportion of defectives is more than 5% less than on the less expensive machine.

Since, the claim of the manufacturer is satisfied. The manufacturer has to buy the more expensive machine.

Thus, the manufacturer has to buy more expensive machine.

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Chapter 9 Solutions

Essential Statistics

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