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:

  1. The sample is obtained using simple random sampling.
  2. The population of piston diameters is normally distributed.

Part 2: Hypotheses for the Test

The null and alternative hypotheses are: H0:σ=0.006H_0: \sigma = 0.006 H1:σ<0.006H_1: \sigma < 0.006 where σ\sigma represents the population standard deviation.


Part 3: Test Statistic Calculation

The test statistic is calculated using the formula: χ2=(n1)s2σ02\chi^2 = \frac{(n-1) \cdot s^2}{\sigma_0^2} where:

  • n=26n = 26 (sample size),
  • s=0.0042s = 0.0042 (sample standard deviation),
  • σ0=0.006\sigma_0 = 0.006 (hypothesized standard deviation).

Substituting values: χ2=(261)(0.0042)2(0.006)2=250.000017640.00003612.25\chi^2 = \frac{(26-1) \cdot (0.0042)^2}{(0.006)^2} = \frac{25 \cdot 0.00001764}{0.000036} \approx 12.25

Thus, the test statistic is: χ2=12.25\chi^2 = 12.25


Part 4: Approximation of the PP-value

To find the PP-value:

  1. The degrees of freedom (df) are n1=261=25n-1 = 26-1 = 25.
  2. Using a chi-square distribution table or calculator, the cumulative probability corresponding to χ2=12.25\chi^2 = 12.25 with 25 degrees of freedom is found.

From the chi-square distribution table, χ2=12.25\chi^2 = 12.25 with 25 degrees of freedom gives a PP-value between 0.05 and 0.10.

The correct answer is: D. The PP-value is between 0.05 and 0.10.


Conclusion

Since the PP-value is not less than the significance level α=0.10\alpha = 0.10, 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:

  1. How is the chi-square distribution used in hypothesis testing?
  2. Why is the null hypothesis stated as σ=0.006\sigma = 0.006 in this test?
  3. How would the conclusion change if the significance level were α=0.05\alpha = 0.05?
  4. How does sample size affect the chi-square test result?
  5. What are the assumptions underlying the chi-square test for variance?

Tip: When interpreting PP-values, always compare it directly with α\alpha 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