Question

Performance Lawn Equipment (PLE) is considering changing their process to make mower blades. They conducted seven production runs and recorded costs. Will the new process save PLE money? Use α = 0.05. (Show all six steps of hypothesis testing.)

Current	New Process
$242.00	$242.00
$176.00	$275.00
$286.00	$199.00
$269.00	$219.00
$327.00	$273.00
$264.00	$265.00
$296.00	$435.00

Answer 1

(1) List all the information

From The Given dta

For Current Year: $\bar{x}$ ₁ = 265.714, s₁ = 47.801, n₁ = 7

For: New Year:     $\bar{x}$ ₂ = 272.571, s₂ = 77.105, n₂ = 7

$\alpha$ = 0.05

(2) State the Null and alternate hypothesis

The Hypothesis:

$H0: \mu_1 = \mu_2:$

$H0: \mu_1 < \mu_2:$

This is a Left tailed test

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(3) State the test to be used and verify the requirements

We use a 2 sample t test, pooled variances, with df = n1 + n2 - 2 = 7 + 7 - 2 = 12

The requirements are

(1) Samples are simple random samples

(2) The samples are independent of each other

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(4) Compute the standardized test statistic

The Pooled Variance:

$S_p^2 = \frac{(n1-1)*s1^2+(n2-1)*s2^2}{n1+n2-2} = \frac{6*47.801^2+6*77.105^2}{7+7-2} = 4115$

The Test Statistic:

$t = \frac{\bar{x}_1-\bar{x}_2}{\sqrt{S_p^2}(\frac{1}{n1}+\frac{1}{n2})} = \frac{266.714-272.572}{\sqrt{4115*(\frac{1}{7}+\frac{1}{7})}} = -0.17$

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(5) Find the p value and compare it to the level of significance.

The p value at t = -0.17) is 0.5665.

p value is > 0.05

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(6) Make a decision, and then write the final conclusion

Fail To Reject H0. There isn't sufficient evidence at the 95% significance level to conclude that new process will save PLE money.

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