Elementary Statistics 2nd Edition
Elementary Statistics 2nd Edition
2nd Edition
ISBN: 9781259724275
Author: William Navidi, Barry Monk
Publisher: McGraw-Hill Education
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Textbook Question
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Chapter 11.4, Problem 13E

Sugar content: A broth used to manufacture a pharmaceutical product has its sugar content, in milligrams per milliliter, measured several times on each of three successive days.

Day 1: 5.0 4.8 5.1 5.1 4.8 5.1 4.8 4.8 5.0 5.2 4.9 4.9 5.0

Day 2: 5.8 4.7 4.7 4.9 5.1 4.9 5.4 5.3 4.8 5.7 5.1 5.7

Day 3: 6.3 4.7 5.1 5.9 5.1 5.9 4.7 6.0 5.3 4.9 5.7 5.3 5.6

  1. Can you conclude that the variability of the process is greater on the second day than on the first day? Use the α = 0.05 level of significance.
  2. Can you conclude that the variability of the process is greater on the third day than on the second day? Use the α = 0.01 level of significance.

(a)

Expert Solution
Check Mark
To determine

To find:

Whether the variability of the process is greater on the second day than on the first day.

Answer to Problem 13E

The variability of the process is greater on the second day than on the first day.

Explanation of Solution

Given Information:

Sugar content: A broth used to manufacture a pharmaceutical product has its sugar content, in milligrams per milliliter, measured several times on each of three successive days.

    Day

      1

      5.04.85.15.14.85.14.84.85.05.24.94.95.0

    Day

      2

      5.84.74.74.95.14.95.45.35.34.85.75.15.7

    Day

      3

      6.34.75.15.95.15.94.76.05.34.95.75.35.6

Formula used:

To test the hypothesis is that the variability of tile process is higher than on second day and first day at 5% level of significance.

First, give hypothesis, compute test-statistics, p-value then give conclusion based on p-value.

The null and alternative hypothesis is.

  H0:σ12=σ22Ha:σ12>σ22

Calculation:

The F-test statistics is.

By using MINITAB, find F-test statistics with the help of following steps is:

  1. Import the data.
  2. Select the Stat and choose the Basic Statistics option
  3. Select the 2Variances and choose variable option and put Sample in different columns
  4. Select Confidence level,
  5. Click 0k

Test and CI for two variances: Day 2 , Day 1 .

Method

  Null hypothesis             Variance(Day 2)/Variance(Day 1)=:Alternative hypothesis   Variance(Day 2)/Variance(Day 1)>:

Statistics

  Variable    N    StDev       VarianceDay 2        13    0.385        0.148Day 1        13    0.139        0.019

Tests

  Method                  TestF Test(normal)DF1DF2     Statistic     p-value7.70                             12       12          7.70          0.001

From the MINITAB output the F test statistics is.

The p-value for this test is,

  0.001 From the MINITAB output, the p-value for this test is.

The conclusion is that the p-value in this context is less than 0.05 which is 0.001 , so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the variability of the process is higher than on second day and first day. The result is statistically significant.

(b)

Expert Solution
Check Mark
To determine

To find:

Whether the vanabihty of the process is greater on the third day than on the second day.

Answer to Problem 13E

The vanabihty of the process is not greater on the third day than on the second day.

Explanation of Solution

Given Information:

To test the hypothesis is that the variability of tile process is higher than on third day and second day at 5% level of significance.

First, give hypothesis, compute test-statistics, p-value then give conclusion based on p-value.

The null and alternative hypothesis is.

  H0:σ22=σ32Ha:σ22>σ32

Formula used:

The F-test statistics is.

By using MINITAB, find F-test statistics with the help of following steps is:

  1. Import the data.
  2. Select the Stat and choose the Basic Statistics option
  3. Select the 2Variances and choose variable option and put Sample in different columns
  4. Select Confidence level,
  5. Click 0k

Calculation:

Test and CI for two variances: Day 3 , Day 2 .

Method

  Null hypothesis             Variance(Day 3)/Variance(Day 2)=1Alternative hypothesis   Variance(Day 3)/Variance(Day 2)>1Significance level           Alpha=0.05

Statistics

  Variable    N    StDev       VarianceDay 3        13    0.518        0.269Day 2        13    0.385        0.148

Tests

  Method                  TestF Test(normal)DF1DF2     Statistic     p-value1.81                             12       12          1.81          0.158

From the MINITAB output the F test statistics is.

The p-value for this test is,

  0.158

The conclusion is that the p-value in this context is higher than 0.05 which is 0.158 , so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the variability of the process is higher than on third day and second day. The result is statistically significant.

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Chapter 11 Solutions

Elementary Statistics 2nd Edition

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