The Rockwell hardness index for steel is determined by pressing a diamond point

Question

The Rockwell hardness index for steel is determined by pressing a diamond point into the steel and measuring the depth of penetration. For 50 specimens of a certain type of steel, the Rockwell index gave a sample mean of 62 with sample standard deviation s-8. The manufacturer claims that this steel has a mean hardness index of 64 a. Using = 0.05 level of significance test whether the mean Rockwell hardness index is less than 64. Please approximate the p-value in order to test the hypothesis b. Based on your answer to part (a), is it still possible that the null hypothesis is true? The steel is sufficiently hard for a certain use so long as the mean Rockwell hardness measure does not drop below 60 C. Suppose that the true mean is 60. If -0.05 what is the probability of detecting the shift from 64 to 60? For this question assume that 8 Suppose that we want the probability of detecting the shift from 64 to 60 to be 99%. Assume that we are willing to take a risk of committing a Type I error = 0.05. For this question assume that d. What sample size should we select?

Explanation / Answer

Here sample size n = 50

sample mean x = 62

Sample standard deviation s = 8

standard error of sample mean se0 = s/sqrt(n) = 8/sqrt(50) = 1.1314

(a) Here Test statisticc

t = (x - 64)/se0 = (62 - 64)/1.1314 = -1.7677

Here dF = 50 - 1 = 49 and alpha = 0.05 and one tailed test so

tcritical = t49,0.05 = 1.6766

p - value = 0.042 < 0.05

so here, l t l > tcritical so we reject the null hypothesis and can claim that mean hardness index is less than 64.

(b) Yes, it is still possivble that null hypothesis is true as there are around 4% probability of that happening, that is called the type I error.

(c) Here population standard deviation = 8

Here we will reject the null hypothesis that true mean is less than 64, if sample mean x < 64 - Z95%* se0 that mean x < 64 - 1.645 * 1.1314 or x < 62.14

so it true population mean = 60

So, Pr(Detecting the shift) = Pr(x< 62.14 ; 60 ; 1.1314)

Z = (62.14 - 60)/1.1314 = 1.89

Pr(Detecting the shift) = Pr(x< 62.14 ; 60 ; 1.1314) = Pr(Z < 1.89) = 0.9707

(d) Now here we want that type II error will be 99%.

so Here that means if sample size is n for that.

we will reject the null hypothesis when x < 64 - 1.645 * 8/n

and as we know the power = 0.99

Pr(x < 64 - 1.645 * 8/n ; 60 ; 8/n) = 0.99

so here Z value for the given probability

Z= 2.326 for p - value 0.99

so,

(64 - 1.645 * 8/n  - 60)/(8/n) = 2.326

4 = (1.645 * 8 + 2.326 * 8)/n

4 = 31.768/n

n = 7.942

n = 63.07 or 64

so sample size required = 64

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