Answer to Question #175551 in Statistics and Probability for Joshua

Question

In a study of distances traveled by buses before the first major engine failure, a sample of 191 buses resulted in a mean of 96,700 miles and a standard deviation of 37,500 miles. At the 0.05 level of signıficance, test the manufacturer's claim that the mean distance traveled before a major engine failure is more than 90,000 miles.

1. Claim:

Ho:

Ha:

2. Level of Significance:

Test- statistic:

Tails in Distribution:

3. Reject Ho if:

4. Compute for the value of the test statistics.

5. Make a decision:

6. State the conclusion in terms of the original problem.

Answer 1

1. Claim:

Ho: µ₁ = µ₀, (the mean distance traveled before a major engine failure is not different from 90,000 miles (µ₀))

Ha: µ₁ > µ₀, (µ₀ =90,000 miles), (the mean distance traveled before a major engine failure is more than 90,000 miles

2. Level of Significance: α=0.05

Test- statistic: Z- statistics, $Z=\frac{\bar{X}-\mu_0}{\frac{s}{\sqrt{n}}}$

$Z=\frac{\bar{X}-\mu_0}{\frac{s}{\sqrt{n}}}$

Tails in Distribution: Right tail or upper-tailed test

3. Reject Ho if: Reject H₀ if $Z\geq 1.645$.

4. Compute for the value of the test statistics.

$Z=$ $\frac{\bar{X}-\mu_0}{\frac{s}{\sqrt{n}}}=\frac{96700-90000}{\frac{37500}{\sqrt{191}}}=2.469$

$\frac{\bar{X}-\mu_0}{\frac{s}{\sqrt{n}}}=\frac{96700-90000}{\frac{37500}{\sqrt{191}}}=2.469$$\frac{\bar{X}-\mu_0}{\frac{s}{\sqrt{n}}}=\frac{96700-90000}{\frac{37500}{\sqrt{191}}}=2.469$

5. Make a decision: We reject H₀ because 2.469 > 1.645.

6. State the conclusion in terms of the original problem.

We have statistically significant evidence at α=0.05, to show that the mean distance traveled before a major engine failure is more than 90,000 miles.