The current rate for producing 5 amp fuses at Neary Electric Co. is 250 per hour

Question

The current rate for producing 5 amp fuses at Neary Electric Co. is 250 per hour. A new machine has been purchased and installed that, according to the supplier, will increase the production rate. The production hours are normally distributed. A sample of 10 randomly selected hours from last month revealed that the mean hourly production on the new machine was 256 units, with a sample standard deviation of 6 per hour. At the.05 significance level can Neary conclude that the new machine is faster?

Explanation / Answer

Below are the null and alternate hypothesis
H0: mu <= 250
H1: mu > 250

alpha = 0.05
sample mean = 256
sample std. dev. = 6
sample size = 10

Test statstics: t (As sample std. dev. is given and sample size is less than 30)

Test statistics t = (xbar - mu)/(sigma/sqrt(n))

t = (256 - 250)/(6/sqrt(10)) = 3.1623

Value from the table that defines the rejection region: 1.65
i.e. Reject H0 if t statistics is greater than 1.65

Decision: reject H0
This means there are significant evidence that the new machine is faster.

Confidence interval formula:
xbar +/- z(alpha/2) * SE
SE = std. dev. / sqrt(n)

SE = 6/sqrt(10) = 1.8974
z(alpha/2) = 1.96

CI = 250 +/- 1.96*1.8974 = (246.2811, 253.7189)

p-value = 0.00575

This is one-tailed test and one sample test

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