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.

Step 1: Formulate Hypotheses

1.     Null Hypothesis (H₀): The mean hardness of the castings is 20.25 (no difference from the general mean).

o    H0:μ=20.25H_0 : \mu = 20.25H0​:μ=20.25

2.     Alternative Hypothesis (H₁): The mean hardness of the castings is greater than 20.25 (supporting the supplier’s claim of higher hardness).

o    H1:μ>20.25H_1 : \mu > 20.25H1​:μ>20.25

This is a one-tailed test since we are specifically interested in whether the mean hardness is higher, not just different.

Step 2: Set Significance Level

Typically, a 5% significance level (α=0.05\alpha = 0.05α=0.05) is used, unless specified otherwise. This means we’ll reject the null hypothesis if the probability of obtaining a sample mean at least as extreme as 20.50 is less than 0.05.

Step 3: Calculate the Test Statistic

The formula for the z-test statistic when testing the mean is:

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

Where:

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

Substitute the values:

z=20.50−20.252.5/100z = \frac{20.50 - 20.25}{2.5 / \sqrt{100}}z=2.5/100​20.50−20.25​ z=0.252.5/10=0.250.25=1z = \frac{0.25}{2.5 / 10} = \frac{0.25}{0.25} = 1z=2.5/100.25​=0.250.25​=1

So, z=1.00z = 1.00z=1.00.

Step 4: Determine the p-Value

For a z-score of 1.00 in a one-tailed test, we check the p-value associated with z = 1.00. Consulting a z-table or standard normal distribution:

p=P(Z>1.00)≈0.1587p = P(Z > 1.00) \approx 0.1587p=P(Z>1.00)≈0.1587

Step 5: Compare p-Value with Significance Level

Since p=0.1587p = 0.1587p=0.1587 is greater than α=0.05\alpha = 0.05α=0.05, we fail to reject the null hypothesis.

Conclusion

There isn’t sufficient evidence to support the supplier’s claim that the mean hardness of the castings is higher than the population mean of 20.25. The observed difference in sample mean (20.50) is not statistically significant at the 5% level, so the claim of "heavier hardness" is not tenable based on this sample data.

 

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