The production process for engine control housing units of a particular type has recently been modified. Prior to this modification, historical data had suggested that the distribution of hole diameters for bushings on the housings was normal with a standard deviation of 0.100 mm. It is believed that the modification has not affected the shape of the distribution or the standard deviation, but that the value of the mean diameter may have changed. A sample of 40 housing units is selected and hole diameter is determined for each one, resulting in a sample mean diameter of 5.426 mm. Let's calculate a confidence interval for true average hole diameter using confidence level of 90%. This requires that 100(1 - a) = , from which a = and z/2 - Z0.05 - 1.645 (corresponding to a cumulative z-curve area of 0.9500). The desired interval is then 0.100 5.426 t = 5.426 + 0.026 = (5.400, 5.452) V40 With a reasonably high degree of confidence, we can say that 5.400 < u < 5.452. This interval is rather narrow because of the small amount of variability in hole diameter (o = 0.100).

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
icon
Related questions
Topic Video
Question
### Understanding Confidence Intervals in Manufacturing

The production process for engine control housing units of a particular type has recently been modified. Initially, it was known that the distribution of hole diameters for bushings on these housings was normal, with a standard deviation of 0.100 mm. After modification, it's assumed that while the distribution shape and standard deviation remain unchanged, the mean diameter might have shifted. A sample of 40 housing units is examined to determine hole diameter, resulting in a sample mean diameter of 5.426 mm.

#### Calculating the Confidence Interval

To find a confidence interval for the true average hole diameter with a confidence level of 90%, we follow these steps:

1. **Determine the Confidence Level**: \(100(1 - \alpha) = 90\%\), thus \(\alpha = 0.10\).

2. **Z-Value**: The z-value for \(\alpha/2\) (0.05) is \(z_{0.05} = 1.645\), which corresponds to a cumulative z-curve area of 0.9500.

3. **Confidence Interval Calculation**:
   \[
   5.426 \pm \left(1.645 \times \frac{0.100}{\sqrt{40}}\right) = 5.426 \pm 0.026 = (5.400, 5.452)
   \]

The desired interval thus ranges from 5.400 mm to 5.452 mm.

#### Conclusion

With reasonably high confidence, it can be said that the true mean diameter \(\mu\) lies within the interval 5.400 mm to 5.452 mm. The interval is narrow, reflecting the small variability in hole diameter (\(\sigma = 0.100\)). This indicates precision in manufacturing despite potential changes in the mean.
Transcribed Image Text:### Understanding Confidence Intervals in Manufacturing The production process for engine control housing units of a particular type has recently been modified. Initially, it was known that the distribution of hole diameters for bushings on these housings was normal, with a standard deviation of 0.100 mm. After modification, it's assumed that while the distribution shape and standard deviation remain unchanged, the mean diameter might have shifted. A sample of 40 housing units is examined to determine hole diameter, resulting in a sample mean diameter of 5.426 mm. #### Calculating the Confidence Interval To find a confidence interval for the true average hole diameter with a confidence level of 90%, we follow these steps: 1. **Determine the Confidence Level**: \(100(1 - \alpha) = 90\%\), thus \(\alpha = 0.10\). 2. **Z-Value**: The z-value for \(\alpha/2\) (0.05) is \(z_{0.05} = 1.645\), which corresponds to a cumulative z-curve area of 0.9500. 3. **Confidence Interval Calculation**: \[ 5.426 \pm \left(1.645 \times \frac{0.100}{\sqrt{40}}\right) = 5.426 \pm 0.026 = (5.400, 5.452) \] The desired interval thus ranges from 5.400 mm to 5.452 mm. #### Conclusion With reasonably high confidence, it can be said that the true mean diameter \(\mu\) lies within the interval 5.400 mm to 5.452 mm. The interval is narrow, reflecting the small variability in hole diameter (\(\sigma = 0.100\)). This indicates precision in manufacturing despite potential changes in the mean.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 3 images

Blurred answer
Knowledge Booster
Hypothesis Tests and Confidence Intervals for Means
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
MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
Elementary Statistics: Picturing the World (7th E…
Elementary Statistics: Picturing the World (7th E…
Statistics
ISBN:
9780134683416
Author:
Ron Larson, Betsy Farber
Publisher:
PEARSON
The Basic Practice of Statistics
The Basic Practice of Statistics
Statistics
ISBN:
9781319042578
Author:
David S. Moore, William I. Notz, Michael A. Fligner
Publisher:
W. H. Freeman
Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
Statistics
ISBN:
9781319013387
Author:
David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:
W. H. Freeman