Math Problem Statement
The piston diameter of a certain hand pump is
0.470.47
inch. The manager determines that the diameters are normally distributed, with a mean of
0.470.47
inch and a standard deviation of
0.0060.006
inch. After recalibrating the production machine, the manager randomly selects
2626
pistons and determines that the standard deviation is
0.00420.0042
inch. Is there significant evidence for the manager to conclude that the standard deviation has decreased at the
alpha equals 0.10α=0.10
level of significance?
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Question content area bottom
Part 1
Are the requirements met to perform this test?
Because the sample
is
obtained using simple random sampling and the population
is
normally distributed, the requirements for testing the hypothesis
are
satisfied.
Part 2
What are the correct hypotheses for this test?
H Subscript 0:
sigmaσ
equals=
0.0060.006
versus
H Subscript 1:
sigmaσ
less than<
0.0060.006
(Type integers or decimals. Do not round.)
Part 3
Calculate the value of the test statistic.
chi Subscript 0 Superscript 2χ20equals=12.2512.25
(Round to two decimal places as needed.)
Part 4
Approximate the P-value for the test statistic. Choose the correct answer below.
A.
The P-value is
greater than 0.10greater than 0.10.
B.
The P-value is
between 0.005 and 0.01between 0.005 and 0.01.
C.
The P-value is
between 0.025 and 0.05between 0.025 and 0.05.
D.
The P-value is
between 0.05 and 0.10between 0.05 and 0.10.
E.
The P-value is
between 0.01 and 0.025between 0.01 and 0.025.
F.
The P-value is
less than 0.005less than 0.005.
Solution
Part 1: Are the requirements met?
Yes, the requirements for performing this test are satisfied because:
- The sample is obtained using simple random sampling.
- The population of piston diameters is normally distributed.
Part 2: Hypotheses for the Test
The null and alternative hypotheses are: where represents the population standard deviation.
Part 3: Test Statistic Calculation
The test statistic is calculated using the formula: where:
- (sample size),
- (sample standard deviation),
- (hypothesized standard deviation).
Substituting values:
Thus, the test statistic is:
Part 4: Approximation of the -value
To find the -value:
- The degrees of freedom (df) are .
- Using a chi-square distribution table or calculator, the cumulative probability corresponding to with 25 degrees of freedom is found.
From the chi-square distribution table, with 25 degrees of freedom gives a -value between 0.05 and 0.10.
The correct answer is: D. The -value is between 0.05 and 0.10.
Conclusion
Since the -value is not less than the significance level , we fail to reject the null hypothesis. There is insufficient evidence to conclude that the population standard deviation has decreased.
Do you have questions or want further clarification?
Here are 5 related questions to explore:
- How is the chi-square distribution used in hypothesis testing?
- Why is the null hypothesis stated as in this test?
- How would the conclusion change if the significance level were ?
- How does sample size affect the chi-square test result?
- What are the assumptions underlying the chi-square test for variance?
Tip: When interpreting -values, always compare it directly with to draw conclusions about statistical significance.
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Math Problem Analysis
Mathematical Concepts
Hypothesis Testing
Chi-Square Distribution
Standard Deviation
Formulas
Chi-square test statistic: χ² = ((n-1) * s²) / σ₀²
Degrees of freedom: df = n - 1
Theorems
Chi-Square Distribution
Properties of Standard Deviation
Suitable Grade Level
Undergraduate Statistics or Advanced High School Statistics
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