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.

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.

Hypothesis Testing for the Supplier's Claim

To test whether the claim of the supplier is tenable, we can use hypothesis testing. The supplier claims that his castings have a heavier hardness, meaning that the mean hardness of his castings is greater than the established mean of 20.25. The sample data from the purchase manager shows a sample mean of 20.50. Our task is to assess whether this observed sample mean is significantly different from the population mean (20.25) to support the supplier's claim.

Step 1: Define the Hypotheses

In hypothesis testing, we start by defining two competing hypotheses:

Null Hypothesis (H₀): The supplier’s claim is not supported. The mean hardness of the castings from the supplier is equal to or less than the established mean of 20.25.
$H_0: \mu = 20.25$
Alternative Hypothesis (H₁): The supplier’s claim is supported. The mean hardness of the castings from the supplier is greater than 20.25 (i.e., the supplier's castings are harder).
$H_1: \mu > 20.25$

This is a one-tailed (right-tailed) test because we are specifically testing if the mean is greater than 20.25.

Step 2: Identify the Test Statistic

Since the population standard deviation (SD) is known (SD = 2.5), we will use the z-test for the sample mean. The z-test is appropriate because we are working with a large sample size (n = 100), which allows us to assume the sampling distribution of the sample mean is approximately normal by the Central Limit Theorem, even though we may not know the population distribution in detail.

The formula for the z-test statistic is:

z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

Where:

$\bar{x}$ = sample mean = 20.50
$\mu$ = population mean (under the null hypothesis) = 20.25
$\sigma$ = population standard deviation = 2.5
$n$ = sample size = 100

Step 3: Calculate the Z-Statistic

Let’s substitute the values into the formula:

z = \frac{20.50 - 20.25}{\frac{2.5}{\sqrt{100}}}

First, calculate the denominator (standard error of the mean):

\frac{2.5}{\sqrt{100}} = \frac{2.5}{10} = 0.25

Now, calculate the z-value:

z = \frac{0.25}{0.25} = 1

Step 4: Determine the Critical Value

To make a decision about the hypothesis, we need to compare the calculated z-value to the critical z-value from the standard normal distribution. Since this is a one-tailed test (we are testing if the mean is greater than 20.25), we look up the critical z-value for a given level of significance (alpha, $\alpha$ ).

Let’s assume a typical significance level of $\alpha = 0.05$ . For a one-tailed test with $\alpha = 0.05$ , the critical z-value is:

z_{\text{critical}} = 1.645

This means that if the calculated z-value is greater than 1.645, we will reject the null hypothesis.

Step 5: Make the Decision

We compare the calculated z-value to the critical z-value:

Calculated z-value = 1.00
Critical z-value = 1.645

Since 1.00 is less than 1.645, we fail to reject the null hypothesis. The sample data does not provide enough evidence to support the claim that the mean hardness of the supplier's castings is greater than 20.25.

Step 6: Conclusion

Based on the hypothesis test, we conclude that there is insufficient evidence to support the supplier's claim that their castings have a mean hardness greater than 20.25. Therefore, the supplier’s claim is not tenable at the 0.05 significance level.

Additional Considerations

Type I and Type II Errors: In this context, a Type I error would occur if we rejected the null hypothesis when it was actually true (i.e., concluding that the supplier’s castings are harder when they are not). A Type II error would occur if we failed to reject the null hypothesis when the alternative hypothesis was true (i.e., concluding that the supplier’s castings are not harder when they actually are).
The probability of a Type I error is set by the significance level $\alpha$ (0.05 in this case), and the probability of a Type II error depends on the true mean hardness, sample size, and the significance level. Since we failed to reject the null hypothesis, there is a chance of committing a Type II error, but based on the sample data and the z-test, we cannot conclude that the supplier’s claim is supported.
Effect Size: The z-value we calculated is relatively small (z = 1.00), which indicates that the observed difference between the sample mean (20.50) and the population mean (20.25) is not large enough to be statistically significant at the 0.05 level. The effect size could be a useful metric to assess the magnitude of the difference, but with this z-value, the effect is not large enough to reject the null hypothesis.
Power of the Test: The power of the test refers to the probability that the test will correctly reject the null hypothesis when the alternative hypothesis is true. In this case, the test had a low power (because the z-value was not large enough), meaning that if the supplier's castings were actually harder than 20.25, the sample size might have been too small to detect this difference. Increasing the sample size would improve the power of the test.
Sample Size Considerations: With a sample size of 100, the test is reasonably well-powered, but if the difference between the sample mean and population mean was very small, a larger sample size might have been needed to detect it. Additionally, larger samples tend to produce more accurate estimates of population parameters, which would help in detecting smaller differences in the mean hardness.

Alternative Tests

While the z-test is suitable here due to the known population standard deviation and large sample size, other tests might be applicable in different scenarios:

If the population standard deviation was unknown, the t-test would be used instead of the z-test. The t-test uses the sample standard deviation and adjusts for the fact that the population variance is unknown, but with large samples, the t-distribution approaches the normal distribution.
If the sample size were smaller or if there were concerns about the normality of the population distribution, other statistical methods, such as bootstrapping or Bayesian methods, could be used to estimate the mean and assess the claim.

In this case, however, the z-test provides a straightforward and appropriate approach to test the hypothesis.

Conclusion

To summarize, using the z-test for hypothesis testing, we found that the supplier’s claim that the hardness of their castings is greater than 20.25 is not statistically supported based on the sample data (mean hardness of 20.50). The calculated z-value of 1.00 is smaller than the critical z-value of 1.645, so we fail to reject the null hypothesis. Thus, there is not enough evidence to support the supplier's claim at the 5% significance level.