Question

The extrusion length of straws follows a normal distribution with a population standard deviation of 0.125 inches. There is concern that the settings for the cut to length encoder has changed from a routing software upgrade. Historically, the length has always been 8 inches. 40 samples were taken at the start of the shift and the average was calculated to be 7.925”. Is there sufficient evidence to suggest that the mean cutting length is different from 8”. (alpha = 0.05)
1. State the Null Hypothesis: a. μ = 7.925 b. μ = 8 c. μ ≠ 8 d. μ ≠ 7.925 2. State the Alternate: a. μ = 7.925 b. μ = 8 c. μ ≠ 8 d. μ ≠ 8.25 3. State the significance level and one or two tailed:

4. State the statistic:

5. Reject Ho if




6. Perform the Calculation – Make the Decision

7. Interpret the result:


8. Extra Credit – Pvalue?

Answer 1

1)

H0 : mu = 8

2)

Ha : mu not = 8

3)

0.05 is the levle of significanc e

this is a two tailed test

4)

z = (xbar -mu)/(sigma/sqrt(n)
= ( 7.925 - 8)/(0.125/sqrt(40))
= -3.79

5)

The z value at 95% confidence interval is,

alpha = 1 - 0.95 = 0.05
alpha/2 = 0.05/2 = 0.025
Zalpha/2 = Z0.025 = 1.96

Critical values are -1.96 and 1.96

reject H0 if z< -1.96 or z > 1.96

6)

Reject H0 as ZStat < -1.96

7)
There is suffiicent evidence to support the claim that the mean cutting length is different from 8

8)

p value = 2 *P(z < -3.79)

P value = 0.0001