Study Tools

Quiz 9.2A

Problem 1: Nonconforming Items in Industrial Process

  • Given Information: 11% of products are known to be nonconforming.

  • Modification Attempt: The company modifies the process to reduce nonconformities.

  • Data from Trial Run: Sample size = 300 items, Nonconforming items = 16.

(a) Test of Significance

  • Hypotheses:

    • Null Hypothesis: ( H_0: p = 0.11 ) (the true proportion of nonconforming items is 11%)

    • Alternative Hypothesis: ( H_a: p < 0.11 ) (the true proportion of nonconforming items is less than 11%)

  • Significance Level: ( \alpha = 0.05 )

  • Procedure: One-sample z-test for a proportion.

    • Conditions:

    • Random: Random sample taken.

    • 10% Condition: Total population assumed to be larger than 3000. (At least ( 10 \times 300 = 3000 ))

    • Large Counts Condition:

      • ( n p_0 = 300 \times 0.11 = 33 ) (greater than 10)

      • ( n (1 - p_0) = 300 \times 0.89 = 267 ) (greater than 10)

  • Sample Proportion:

    • Calculate: ( \hat{p} = \frac{16}{300} = 0.0533 )

  • Z-Score Calculation:

    • [ z = \frac{\hat{p} - p0}{\sqrt{\frac{p0(1 - p_0)}{n}}} ]

    • Calculate: [ z = \frac{0.0533 - 0.11}{\sqrt{\frac{0.11 \times 0.89}{300}}} = -3.14 ]

  • P-value:

    • Using z-table, P-value for z = -3.14 is approximately 0.0008.

  • Conclusion:

    • Since P-value (0.0008) < α (0.05), reject ( H_0 ).

    • Evidence: Convincing evidence that true proportion of nonconforming items is less than 0.11.

(b) Explanation of P-value

  • Interpretation: If ( H_0 ) (true proportion = 0.11) is correct, there is a 0.0008 probability of obtaining a sample proportion that is as far or further below 0.11 as 0.0533.

Problem 2: Germination Rate of Seeds

  • Context: A grass seed variety with a typical germination rate of 0.80.

  • Objective: Company sprays seeds with a chemistry known to change germination rates for testing.

(a) Power of the Test

  • Sample Information: Sample size = 400 seeds, Two-sided test for ( H_0: p = 0.80 ), significance level = ( \alpha = 0.05 ).

  • Power Definition: Power = 0.69, which measures the probability of correctly rejecting ( H_0 ) when the true proportion is ( p = 0.75 ).

(b) Increasing Power of the Test

  • Methods: Two ways to increase the power of the test:

    • Increase Sample Size: Larger sample size increases the likelihood of detecting an effect if there is one.

    • Increase Significance Level: Raising ( \alpha ) increases the probability of rejection of the null hypothesis.

(c) Confidence Interval Analysis

  • Germination Results: 307 out of 400 seeds germinate.

  • 95% Confidence Interval: ( (0.726, 0.809) ) for the proportion of seeds that germinate.

  • Conclusion:

    • Since the confidence interval contains the null value of 0.80, we fail to reject ( H_0 ).

    • Evidence: Not convincing that the germination rate changed due to the chemical spray.

Quiz 9.2B

Problem 1: Proportion of Red Cars at Citizens Bank Park

  • Known Proportion: 0.12 of cars nationally are red.

  • Data Collection: SRS of 210 cars parked, 35 counted red.

(a) Test of Significance

  • Hypotheses:

    • Null Hypothesis: ( H_0: p = 0.12 )

    • Alternative Hypothesis: ( H_a: p > 0.12 )

  • Significance Level: ( \alpha = 0.05 )

  • Procedure: One-sample z-test for proportion.

    • Conditions:

    • Random: SRS of 210 cars taken.

    • 10% Condition: 210 is less than 10% of 21,000 car total (allowed).

    • Large Counts Condition:

      • ( np = 210 \times 0.12 = 25.2 ) (greater than 10)

      • ( n(1 - p) = 210 \times 0.88 = 184.8 ) (greater than 10)

  • Sample Proportion:

    • Calculate: ( \hat{p} = \frac{35}{210} = 0.1667 )

  • Z-Score Calculation:

    • [ z = \frac{0.1667 - 0.12}{\sqrt{\frac{0.12 \times 0.88}{210}}} ]

    • Calculate: [ z = \frac{0.0467}{0.0335} = 2.081 ]

  • P-value:

    • From z-table, P-value for z = 2.081 is approximately 0.0187.

  • Conclusion:

    • Since P-value (0.0187) < α (0.05), reject ( H_0 ).

    • Evidence: Convincing evidence that the true proportion of red cars at Citizens Bank Park is greater than 0.12.

(b) Explanation of P-value

  • Interpretation: If ( H_0 ) (true proportion = 0.12) is correct, there is a 0.0187 probability of obtaining a sample proportion as far or further above 0.12 as 0.1667.

Problem 2: Impact of Attack Ads on Candidate Support

  • Context: Candidate supported by 44% of voters.

  • Objective: Assess if support changed after an attack advertisement.

(a) Power of the Test

  • Hypotheses:

    • Null Hypothesis: ( H_0: p = 0.44 )

    • Alternative Hypothesis: ( H_a: p < 0.44 )

  • Sample Size: 450 voters surveyed.

  • Power Definition: Power = 0.36, probability of correctly rejecting ( H_0 ) when the true proportion is ( p = 0.40 ).

(b) Increasing Power of the Test

  • Methods: Two ways to increase the power of this test:

    • Increase the Sample Size: To reduce sampling error, which increases power.

    • Increase the Significance Level: Raising ( \alpha ) increases chances of rejection of ( H_0 ).

(c) Confidence Interval Analysis

  • Survey Findings: Out of 450 voters, 186 support the candidate.

  • 95% Confidence Interval for Support: ( (0.368, 0.459) )

  • Conclusion:

    • Since the confidence interval includes null value of 0.44, we fail to reject ( H_0 ).

    • Evidence: No convincing evidence that the attack ads changed support levels.