A purchase manager knows that the hardness of castings from any supplier is normally distributed with a mean of 20.25 and SD of 2.5. He picks up 100 samples of castings from any supplier who claims that his castings have heavier hardness and finds the mean hardness as 20.50. Test whether the claim of the supplier is tenable.

Posted By: December 15, 2024 Leave a Reply

Q. A purchase manager knows that the hardness of castings from any supplier is normally distributed with a mean of 20.25 and SD of 2.5. He picks up 100 samples of castings from any supplier who claims that his castings have heavier hardness and finds the mean hardness as 20.50. Test whether the claim of the supplier is tenable.

1. Problem Statement and Objective

The purchase manager is tasked with evaluating whether a supplier's claim that the hardness of their castings is higher than the historical average is valid. The historical data shows that the hardness of castings from any supplier is normally distributed with a mean (μ) of 20.25 and a standard deviation (σ) of 2.5. A sample of 100 castings from the supplier in question has been selected, and the sample mean hardness (x̄) is found to be 20.50. The objective of the analysis is to test whether the supplier's claim, that their castings have a higher average hardness, is statistically valid.



2. Formulating Hypotheses

The first step in hypothesis testing is to clearly define the null hypothesis (H₀) and the alternative hypothesis (H₁). In this case, the hypotheses can be formulated as follows:

·        Null Hypothesis (H₀): The supplier's castings have the same average hardness as the historical mean. Mathematically, this is stated as:

H0:μ=20.25H_0: \mu = 20.25H0​:μ=20.25

·        Alternative Hypothesis (H₁): The supplier's castings have a higher average hardness than the historical mean. This is a one-tailed test, meaning we are only interested in whether the new mean is greater than the historical mean. Mathematically, this is stated as:

H1:μ>20.25H_1: \mu > 20.25H1​:μ>20.25

This is a one-tailed hypothesis because the claim from the supplier suggests that the mean hardness is greater than 20.25, and we are testing this specific direction of difference.

3. Significance Level

Next, we need to set the significance level (α), which represents the probability of rejecting the null hypothesis when it is actually true. The typical value for α in hypothesis testing is 0.05, which corresponds to a 5% risk of making a Type I error (i.e., rejecting the null hypothesis when it is true).

Thus, we will use α = 0.05 for this test.

4. Test Statistic

Since the hardness of castings is normally distributed, we can use a z-test to test the hypothesis. The z-test is appropriate because the population standard deviation (σ) is known, and the sample size (n) is large enough (n = 100) to invoke the Central Limit Theorem (which ensures that the sampling distribution of the sample mean is approximately normal).

The formula for the z-statistic is given by:

z=xˉ−μσnz = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}z=n​σ​xˉ−μ​

Where:

  • xˉ\bar{x}xˉ is the sample mean (20.50),
  • μ\muμ is the population mean (20.25),
  • σ\sigmaσ is the population standard deviation (2.5),
  • nnn is the sample size (100).

    Substituting the values into the formula:

    z=20.50−20.252.5100=0.252.510=0.250.25=1.00z = \frac{20.50 - 20.25}{\frac{2.5}{\sqrt{100}}} = \frac{0.25}{\frac{2.5}{10}} = \frac{0.25}{0.25} = 1.00z=100​2.5​20.50−20.25​=102.5​0.25​=0.250.25​=1.00

    Thus, the calculated z-value is 1.00.

    5. Critical Value and Decision Rule

    To make a decision, we need to compare the calculated z-value with the critical z-value for a one-tailed test at the 0.05 significance level. Using standard z-tables or statistical software, the critical z-value for a one-tailed test with α = 0.05 is 1.645.

    • If the calculated z-value is greater than or equal to 1.645, we reject the null hypothesis.
    • If the calculated z-value is less than 1.645, we fail to reject the null hypothesis.

    In our case, the calculated z-value is 1.00, which is less than the critical value of 1.645.

    6. Conclusion

    Since the calculated z-value (1.00) is less than the critical value (1.645), we fail to reject the null hypothesis. This means there is insufficient evidence to support the supplier's claim that their castings have a higher average hardness than the historical mean of 20.25.

    7. Interpretation of Results

    Failing to reject the null hypothesis does not mean that the supplier’s claim is false; it simply means that based on the sample data, there is not enough statistical evidence to prove that their castings are significantly harder than the historical average. The sample mean of 20.50 is only slightly higher than the population mean of 20.25, and the difference is not large enough to be considered statistically significant at the 0.05 level.

    8. Type I and Type II Errors

    It is also important to consider the potential for Type I and Type II errors in hypothesis testing:

    • Type I Error (False Positive): This occurs if we reject the null hypothesis when it is actually true. In this case, a Type I error would occur if we concluded that the supplier’s castings are significantly harder than the historical mean when, in reality, they are not.
    • Type II Error (False Negative): This occurs if we fail to reject the null hypothesis when it is actually false. In this case, a Type II error would occur if the supplier’s castings were truly harder than the historical mean, but we failed to detect the difference because the sample size was too small or the observed difference was too small.

    Given the sample size of 100, the risk of a Type II error is relatively low, but it is always a factor to consider in hypothesis testing.

    9. Possible Reasons for the Outcome

    There are several factors that could explain why the hypothesis test did not show significant evidence for the supplier’s claim:

    • Small Difference Between Sample Mean and Population Mean: The difference between the sample mean (20.50) and the population mean (20.25) is small (only 0.25). This difference may not be large enough to be detected as statistically significant with the sample size of 100, especially given that the population standard deviation is 2.5.
    • Sample Variability: While the sample mean is 20.50, the individual castings within the sample may still exhibit considerable variability. This variability can affect the power of the test to detect small differences.
    • Sample Size and Power: Although 100 samples is generally considered a sufficiently large sample size for many statistical tests, it is always worth considering whether the sample size is large enough to detect a meaningful difference. A larger sample size could potentially increase the power of the test.

    10. Recommendation for the Purchase Manager

    Given the results of the hypothesis test, the purchase manager should be cautious in accepting the supplier’s claim that their castings have a higher average hardness. While the claim is not statistically supported at the 5% significance level, it might still be worth conducting further analysis or gathering additional data.

    • Additional Testing: The purchase manager could consider taking a larger sample from the supplier to increase the power of the hypothesis test. A larger sample size would reduce the standard error, making it easier to detect smaller differences.
    • Exploring Other Factors: The purchase manager could also investigate other factors that may influence the hardness of the castings, such as manufacturing processes, materials used, or environmental conditions.
    • Supplier Communication: It might also be useful to discuss the findings with the supplier to understand if there are any variations in the manufacturing process that could explain the observed differences in sample hardness.

    11. Further Statistical Considerations

    In addition to the basic hypothesis test, the purchase manager could consider performing a confidence interval estimation for the population mean to gain a range of values within which the true population mean is likely to fall. A 95% confidence interval for the population mean can be calculated as follows:

    CI=xˉ±zα/2×σnCI = \bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}CI=xˉ±zα/2​×n​σ​

    Substituting the values:

    CI=20.50±1.96×2.5100=20.50±1.96×0.25=20.50±0.49CI = 20.50 \pm 1.96 \times \frac{2.5}{\sqrt{100}} = 20.50 \pm 1.96 \times 0.25 = 20.50 \pm 0.49CI=20.50±1.96×100​2.5​=20.50±1.96×0.25=20.50±0.49

    Thus, the 95% confidence interval for the population mean is:

    [20.01,20.99][20.01, 20.99][20.01,20.99]

    This interval suggests that, with 95% confidence, the true population mean hardness is between 20.01 and 20.99. Since the historical mean of 20.25 falls within this interval, it further suggests that the sample data does not provide strong evidence of a higher mean.

    12. Final Thoughts

    In conclusion, while the sample mean hardness of 20.50 is slightly higher than the historical mean of 20.25, the hypothesis test does not provide strong enough evidence to support the supplier’s claim of a higher average hardness. The sample size and the observed difference in means are not sufficient to reject the null hypothesis at the 0.05 significance level. Therefore, the purchase manager should remain cautious about accepting the supplier’s claim without further investigation or additional data

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