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Case Problem Air Force Training Program An Air Force introductory course in elec

ID: 2946827 • Letter: C

Question

Case Problem Air Force Training Program An Air Force introductory course in electronics uses a personalized system of instruction whereby instruction text. The students work independently with the text until they have completed the training and passed a test. Of concern is the varying pace at which the students complete this portion of their training program. Some students are able to cover the programmed instruction text relatively quickly, whereas other students work much longer with the text and require additional time to complete the course. The fast students wait until the slow students complete the introductory course before the entire group proceeds together with other aspects of their training each student views a videotaped lecture and then is given a programmed A proposed alternative system involves use of computer-assisted instruction. I method, all students view the same videotaped lecture and then each is assigned to a computer terminal for further instruction. The computer guides the student, working independently, through the self-training portion of the course onterino class of 122 o comare the e pronosed and current methods of instauction

Explanation / Answer

data given:

1.)

Checking mean:

mean(current): 75.06

mean(computer): 75.42

Let's perform t test to find if the difference in mean is significant at alpha = 0.01

Null Hypotheses, Ho: There is no significant difference between mean of 2 populations

Ha: The mean of 2 populations are significantly different

Difference Scores Calculations

Treatment 1

N1: 61
df1 = N - 1 = 61 - 1 = 60
M1: 75.07
SS1: 933.74
s21 = SS1/(N - 1) = 933.74/(61-1) = 15.56


Treatment 2

N2: 61
df2 = N - 1 = 61 - 1 = 60
M2: 75.43
SS2: 376.92
s22 = SS2/(N - 1) = 376.92/(61-1) = 6.28


T-value Calculation

s2p = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22) = ((60/120) * 15.56) + ((60/120) * 6.28) = 10.92

s2M1 = s2p/N1 = 10.92/61 = 0.18
s2M2 = s2p/N2 = 10.92/61 = 0.18

t = (M1 - M2)/?(s2M1 + s2M2) = -0.36/?0.36 = -0.6

The t-value is -0.60268. The p-value is .273928. The result is not significant at p < .01

So, we can reject the alternate hypotheses and with 99.9% confidence we can say that mean of 2 population based on test performed on above 2 samples is significantly same.

3.)

Current Computer

Variance 15.56 6.28

Standard deviation 3.94 2.50

To test equality of population variances, we will perform F-Test.

We will be checking for upper one-tailed test since we can see that there is good amount of difference in variance of 2 samples.

So, for alpha = 0.01, Hypothesis becomes,

Ho: variance(current) = variance(computer)

Ha: variance(current) > variance(computer)

for 60 degrees of freedom and alpha = 0.01

critical F value = 1.836

(3) Test Statistics

The F-statistic is computed as follows:

F = 15.56 / 6.28 = 2.478

Since from the sample information we get that F = 2.478 > F_U = 1.836F=2.478>F-critical, it is then concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population variance sigma_1^2?12? is greater than the population variance sigma_2^2?22?, at the lpha = 0.01?=0.01 significance level.

4.) Looks like mean of Population for noth samples is not different significantly and is almost same.

But from F test it can be clealrly said that there is more variability in time taken by Population with current method.

But with the other population who is using computer, the variability is very less.

So, it can be related to adaptability of students as some students are easilt adaptable to current method ans some are not ansd so resulting in variablity.

On the other hand, in computer based method, all students are taking almost same time. So we can say that all students are at same level in this one.

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