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3 years ago

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Using traditional methods it takes 92 hours to receive an advanced flying license. A new training technique using Computer Aided

Instruction (CAI) has been proposed. A researcher used the technique on 70 students and observed that they had a mean of 93 hours. Assume the population variance is known to be 36. Is there evidence at the 0.05 level that the technique lengthens the training time?
Find the value of the test statistic. Round your answer to two decimal places.

1 answer:

DanielleElmas [232]3 years ago

8 0

Answer:

The value of the test statistic is z = 1.39.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

Step-by-step explanation:

Using traditional methods it takes 92 hours to receive an advanced flying license.

This means that at the null hypothesis, it is tested if the mean is of 92, that is:

$H_0: \mu = 92$

Test if there is evidence that the technique lengthens the training time

At the alternative hypothesis, it is tested if the mean is more than 92, that is:

$H_1: \mu > 92$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

92 is tested at the null hypothesis:

This means that $\mu = 92$

A researcher used the technique on 70 students and observed that they had a mean of 93 hours. Assume the population variance is known to be 36.

This means that $n = 70, X = 93, \sigma = \sqrt{36} = 6$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{93 - 92}{\frac{6}{\sqrt{70}}}$

$z = 1.39$

The value of the test statistic is z = 1.39.

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 93, which is 1 subtracted by the p-value of z = 1.39.

Looking at the z-table, z = 1.39 has a p-value of 0.9177.

1 - 0.9177 = 0.0823.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

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Step-by-step explanation:

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Plot the vertices of the polygon to better understand the problem

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using a graphing tool

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