Question

2 (30 points). "Roller straightening" is a s er struightening" is a method used to straightening wire before coiling it to make a spring, Suppose that a sample of lé wires is randomly selected and each is lested to determin standard deviation are 2162 and 31, respectively e tensile strength (Nmm). The resulting sample mean and sample Assuming the tensile strength distribustion is approximately normal, test Ho: μ-2150 vs Her> 2150, using a-0.05 (b) Calculate the P-valac of the above test. (c) Determine p2175).

Answer 1

Right Tailed t test, Single Mean

(a) Given: $\mu$ = 2150 N/mm², $\bar{x}$ = 2162 N/mm², s = 31 N/mm², n = 16, $\alpha$ = 0.05

The Hypothesis:

H0: $\mu$ = 2150

Ha: $\mu$ > 2150

This is a Right tailed test

The Test Statistic: Since the population standard deviation is unknown, we use the students t test.

The test statistic is given by the equation:

$t = \frac{\bar{x}-\mu }{\frac{s}{\sqrt{n}}} = \frac{2162-2150}{\frac{31}{\sqrt{16}}} = 1.548$

t observed = 1.548

The Critical Value: Right Tailed at $\alpha$ = 0.05, df = 15 is 1.753

The Decision: Since t observed is < tcritical, we Fail to reject H0

The Conclusion: There is insufficient evidence at $\alpha$ = 0.05 to conclude that the mean tensile strength is greater than 2150.

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(b) The p Value: The p value (Right tailed) for t = 1.548, for degrees of freedom (df) = n-1 = 15, is; p value = 0.0712

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(c) To find the probability of a Type II error $\beta$ , when the true mean is 2175

Hypothesized mean = 2150, n = 16, tcritical = 1.753

The value of $\bar{x}$ , for which H0 gets rejected: 1.753 = ( $\bar{x}$ - 2150) / [31/sqrt(16)]

Solving we get, (1.753 * 31 / 4) + 2150 = 2163.59

P(X > 2163.59), when using the hypothesized mean = (2163.59 - 2175)/[31/sqrt(16)] = -1.47

The p value(right tail), df = 15, $\beta$ = 0.9188

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