Question

Using traditional methods it takes 105 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 210 students and observed that they had a mean of 106 hours. Assume the population standard deviation is known to be 6. Is there evidence at the 0.05 level that the technique performs differently than the traditional method?

step 1 of 5 : State the null and alternative hypotheses.

Step 2 of 5 :  Find the value of the test statistic. Round your answer to two decimal places.

Step 3 of 5 : Specify if it is one tailed or two tailed

Step 4 of 5 :  Find the P-value of the test statistic. Round your answer to four decimal places.

Step 5 of 5 : Make the decision to reject or fail to reject the null hypothesis.

Answer 1

Solution:

Given:

the population standard deviation =

Sample size = n = 210

Sample mean =

Step 1) State H0 and H1:

Using traditional methods it takes 105 hours to receive an advanced flying license.

We have to test if there is evidence at the 0.05 level that the technique performs differently than the traditional method.

Since this statement is non-directional, this is two tailed test.

Thus

Vs

Step 2)  Find the value of the test statistic.

Step 3) Specify if it is one tailed or two tailed

This is two tailed test.

Step 4) Find the P-value of the test statistic.

P-value = 2 X P( Z > z)

P-value = 2 X P( Z > 2.42 )

P-value = 2 X [ 1 - P( Z < 2.42 ) ]

Look in z table for z = 2.4 and 0.02 and find corresponding area.

P( Z< 2.42 ) = 0.9922

thus

P-value = 2 X [ 1 - P( Z < 2.42 ) ]

P-value = 2 X [ 1 - 0.9922 ]

P-value = 2 X 0.0078

P-value = 0.0156

Step 5) Make the decision to reject or fail to reject the null hypothesis.

Since P-value = 0.0156 < 0.05 level of significance , we reject null hypothesis.