7q1r2defects250131311115014151413122213121023100211110630 A quality control engineer at a particular Icd screen manufacturer is studying the mean number of defects per screen.Based on historical evidence, the mean number of defects per screen was thought to be 2.58. There have recently beenchanges to the manufacturing process, and the engineer now feels that the mean number of defects per screen may besignificantly smaller than 2.58. Using the number of defects on each of 50 sample screens shown below, conduct theappropriate hypothesis test using a 0.1 level of significance.Assignment 7q1 dataa) What are the appropriate null and alternative hypotheses?2.58 versus Ha: u < 2.58Ho:Но: х= 2.58 versus Ha: x > 2.58Но: и= 2.58 versus Ha: u # 2.58Но: и %3D 2.58 versus Ha: И> 2.58b) What is the test statistic? Give your answer to four decimal places.c) What is the P-value for the test? Give your answer to four decimal places.d) What is the appropriate conclusion?Reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1.Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1.Reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1.Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1

Question
Asked Oct 30, 2019
18 views
7q1r2
defects
2
5
0
1
3
1
3
1
1
1
1
5
0
1
4
1
5
1
4
1
3
1
2
2
2
1
3
1
2
1
0
2
3
1
0
0
2
1
1
1
1
0
6
3
0
help_outline

Image Transcriptionclose

7q1r2 defects 2 5 0 1 3 1 3 1 1 1 1 5 0 1 4 1 5 1 4 1 3 1 2 2 2 1 3 1 2 1 0 2 3 1 0 0 2 1 1 1 1 0 6 3 0

fullscreen
A quality control engineer at a particular Icd screen manufacturer is studying the mean number of defects per screen.
Based on historical evidence, the mean number of defects per screen was thought to be 2.58. There have recently been
changes to the manufacturing process, and the engineer now feels that the mean number of defects per screen may be
significantly smaller than 2.58. Using the number of defects on each of 50 sample screens shown below, conduct the
appropriate hypothesis test using a 0.1 level of significance.
Assignment 7q1 data
a) What are the appropriate null and alternative hypotheses?
2.58 versus Ha: u < 2.58
Ho:
Но: х
= 2.58 versus Ha: x > 2.58
Но: и
= 2.58 versus Ha: u # 2.58
Но: и %3D 2.58 versus Ha: И> 2.58
b) What is the test statistic? Give your answer to four decimal places.
c) What is the P-value for the test? Give your answer to four decimal places.
d) What is the appropriate conclusion?
Reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1.
Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1.
Reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1.
Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1
help_outline

Image Transcriptionclose

A quality control engineer at a particular Icd screen manufacturer is studying the mean number of defects per screen. Based on historical evidence, the mean number of defects per screen was thought to be 2.58. There have recently been changes to the manufacturing process, and the engineer now feels that the mean number of defects per screen may be significantly smaller than 2.58. Using the number of defects on each of 50 sample screens shown below, conduct the appropriate hypothesis test using a 0.1 level of significance. Assignment 7q1 data a) What are the appropriate null and alternative hypotheses? 2.58 versus Ha: u < 2.58 Ho: Но: х = 2.58 versus Ha: x > 2.58 Но: и = 2.58 versus Ha: u # 2.58 Но: и %3D 2.58 versus Ha: И> 2.58 b) What is the test statistic? Give your answer to four decimal places. c) What is the P-value for the test? Give your answer to four decimal places. d) What is the appropriate conclusion? Reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1. Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1. Reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1. Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1

fullscreen
check_circle

Expert Answer

Step 1

a)

The hypotheses for the test are stated below:

From the information, the level of significance is 0.10 and the claim of the test is the mean number of defects per screen may be significantly smaller than 2.58.

The claim of the test is there has been no change in the mean length.

Null hypothesis:

H0: µ = 2.58

Alternative hypothesis:

Ha: µ < 2.58

The correct answer is H0: µ = 2.58 versus Ha: µ < 2.58.

b)

The test statistics is obtained below:

Instructions to obtain the p-value for t test are listed:

  1. In EXCEL, Select Add-Ins > PHStat > One-Sample Tests > t Test for the mean, Sigma unknown.
  2. Enter 58 under Null hypothesis.
  3. Enter 10 under Level of significance.
  4. In Sample Statistic Options, choose Sample Statistic unknown $A1$A51.
  5. In Test Options, choose Two (or Upper or Lower) Tail Test.
  6. In Output Options, enter a Title and click OK.

Follow the above instructions to obtain the below output:

help_outline

Image Transcriptionclose

t Test for Hypothesis of the Mean Data Null Hypothesis 2.58 Level of Significance 0.1 Sample Size Sample Mean Sample Standard Deviation 50 1.86 1.525430685 Intermediate Calculations 0.2157 Standard Error of the Mean Degrees of Freedom 49 t Test Statistic -3.3375 Lower-Tail Test -1.2991 Lower Critical Value p-Value 0.0008 Reject the null hypothesis

fullscreen
Step 2

From MegaStat output, the test statistics is ­−3.3375.

Therefore, the test statistics is −3.3375.

c)

From MegaStat output, the ...

Want to see the full answer?

See Solution

Check out a sample Q&A here.

Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in

Math

Statistics

Related Statistics Q&A

Find answers to questions asked by student like you
Show more Q&A
add
question_answer

Q: for a normal distribution with a mean of 7 and a stanard deviation of 6, the value 10 has a z value ...

A: It is given that mean and standard deviation are 7 and 6, respectively.

question_answer

Q: Question Seventy cards are numbered 1 through 70, one number per card. One card is randomly selec...

A: Define the given events A as multiples of 3 and B as multiples of 5.The events A and B are as follow...

question_answer

Q: Births are approximately uniformly distributed between the 52 weeks of the year. 1 1 sxs53 f(x) = 52

A: Solution:Continuous uniform distribution:A random variable X is said to have the rectangular distrib...

question_answer

Q: 1) For a two-tailed hypothesis using a z-distribution, find the critical values (z-scores)that will ...

A: (a)Given: alpha = 0.20.For two tailed test, critical Z value can be calculated as:

question_answer

Q: An automobile manufacturer finds that 1 in every 2000 automobiles produce has a particular manufactu...

A: Binomial distribution:The events that has exactly two possible outcomes (success or failure) follows...

question_answer

Q: explain please Discrete random variables (rv); definition, distributions, calculation of probabiliti...

A: Hello there! there are more than 3 sub parts in the question. According to our policies cannot answe...

question_answer

Q: Question Help An investment counselor calls with a hot stock tip. He believes that if the economy re...

A: Expected value:The expected value of a random variable is the sum of the possible values that the ra...

question_answer

Q: What would the 95% confidence limits for a sample of 120 with a mean of 51 and a standard deviation ...

A: We are given: 

question_answer

Q: Suppose an individual was studying temperature’s effect on the crime rate. The researcher looked at ...

A: Solution:The sample mean is obtained below:From the given information, number of observations is 6 a...