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Using traditional methods it takes 11.4 hours to receive a basic flying license. A new license training method using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique on 23 students and observed that they had a mean of 11.0 hours with a variance of 3.61. Is there evidence at the 0.025 level that the technique reduces the training time? Assume the population distribution is approximately normal. Step 2 of 5: Find the value of the test statistic. Round your answer to three decimal places.

Sagot :

Answer:

The value of the test statistic is [tex]t = -1.01[/tex]

The pvalue of the test is of 0.1617 > 0.025, which means that there is no evidence at the 0.025 level that the technique reduces the training time.

Step-by-step explanation:

Using traditional methods it takes 11.4 hours to receive a basic flying license.  Is there evidence at the 0.025 level that the technique reduces the training time?

This means that at the null hypothesis, we test that the mean is of 11.4 hours, that is:

[tex]H_0: \mu = 11.4[/tex]

At the alternate hypothesis, we test that the mean is less than this, that is:

[tex]H_a: \mu < 11.4[/tex]

The test statistic is:

[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]s[/tex] is the standard deviation and n is the size of the sample.

11.4 is tested at the null hypothesis:

This means that [tex]\mu = 11.4[/tex]

A researcher used the technique on 23 students and observed that they had a mean of 11.0 hours with a variance of 3.61.

This means that [tex]n = 23, X = 11, s = \sqrt{3.61} = 1.9[/tex]

Value of the test-statistic:

[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{11 - 11.4}{\frac{1.9}{\sqrt{23}}}[/tex]

[tex]t = -1.01[/tex]

Pvalue of the test and decision:

The pvalue of the test is the probability of finding a sample mean of 11 hours or less, which is the pvalue of t = -1.01 with 23 - 1 = 22 degrees of freedom.

Using a calculator, this pvalue is of 0.1617.

The pvalue of the test is of 0.1617 > 0.025, which means that there is no evidence at the 0.025 level that the technique reduces the training time.

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