STAT TECH IN BUSINESS & ECON AC
STAT TECH IN BUSINESS & ECON AC
18th Edition
ISBN: 9781264731657
Author: Lind
Publisher: MCG
bartleby

Videos

Textbook Question
Book Icon
Chapter 12, Problem 1SR

Steele Electric Products Inc. assembles cell phones. For the last 10 days, Mark Nagy completed a mean of 39 phones per day, with a standard deviation of 2 per day. Debbie Richmond completed a mean of 38.5 phones per day, with a standard deviation of 1.5 per day. At the .05 significance level, can we conclude that there is more variation in Mark’s daily production?

Expert Solution & Answer
Check Mark
To determine

Find whether there is more variation in Person M’s daily production.

Answer to Problem 1SR

There is no more variation in Person M’s daily production.

Explanation of Solution

Here, σ12 represents the variation in Person M’s daily production and σ22 represents the variation in Person R’s daily production.

The null and alternative hypotheses are stated below:

H0:σ12σ22

That is, the variation in Person M’s daily production is less than or equal to the variation in Person R’s production.

H1:σ12>σ22

That is, the variation in Person M’s daily production is more than that of Person R’s production.

Step-by-step procedure to obtain the test statistic using MINITAB software:

  • Choose Stat > Basic Statistics > 2 Variance.
  • Under Data, choose Sample standard deviation.
  • In First, enter 10 under Sample size.
  • In First, enter 2 under Standard deviation
  • In Second, enter 10 under Sample size.
  • In Second, enter 1.5 under Standard deviation
  • Check Options, enter Confidence level as 95.0.
  • In Hypothesized ratio StDev 1 / StDev 2
  • Choose greater than in alternative.
  • Click OK in all dialog boxes.
  • Output obtained using MINITAB is represented as follows:

STAT TECH IN BUSINESS & ECON AC, Chapter 12, Problem 1SR

  • From the above output, the F test statistic value is 1.78 and the p-value is 0.202.

Decision Rule:

If the p-value is less than the level of significance, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.

Conclusion:

The level of significance is 0.05.

From the output, the p-value is 0.202.

The p-value is greater than the level of significance 0.05. Hence, one is failed to reject the null hypothesis at the 0.05 significance level.

Therefore, there is no more variation in Person M’s daily production.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
17. Suppose that X1, X2,..., Xn are random variables, such that E|xk| < ∞ for all k, and set Yn = max1
6. Show that, for any random variable, X, and a > 0, L P(x < X ≤ x+a) dx = a. 2015
15. This problem extends Problem 20.6. Let X, Y be random variables with finite mean. Show that (P(X ≤ x ≤ Y) - P(Y < x ≤ X))dx = E Y — E X.

Chapter 12 Solutions

STAT TECH IN BUSINESS & ECON AC

Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Hypothesis Testing using Confidence Interval Approach; Author: BUM2413 Applied Statistics UMP;https://www.youtube.com/watch?v=Hq1l3e9pLyY;License: Standard YouTube License, CC-BY
Hypothesis Testing - Difference of Two Means - Student's -Distribution & Normal Distribution; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=UcZwyzwWU7o;License: Standard Youtube License