Problem 1. We are interested in comparing the efficiencies of two teaching metho
ID: 3273898 • Letter: P
Question
Problem 1. We are interested in comparing the efficiencies of two teaching methods. Six students were taught using each method and their performance scores are shown in the following table.
Method 1
Method 2
79
71
66
43
57
58
91
78
42
20
59
56
Assume that the twelve students are different. Perform the appropriate analysis regarding the comparison of the efficiencies of two teaching methods by using only descriptive statistics in R (i.e. mean() and var()). In particular, state the null and alternative hypothesis, conduct the test at alpha level 0.01, derive a symmetric 99% confidence interval for the parameters of interest. Give interpretations for all relevant quantities.
Perform the analysis from part a) in R and confirm your results.
Assume that the same six students were tested twice after being taught each method. Perform the appropriate analysis regarding the comparison of the efficiencies of two teaching methods by using only descriptive statistics in R. In particular, state the null and alternative hypothesis, conduct the test at alpha level 0.01, derive a symmetric 99% confidence interval for the parameters of interest. Give interpretations for all relevant quantities.
Perform the analysis from part c) in R and confirm your results.
Test the hypothesis that the average performance score for students taught using Method 1 is 80. Construct a symmetric 90% confidence interval for the population average score.
Test the hypothesis that the variance of the performance scores for students taught using Method 1 is 100. Construct a 95% confidence interval for the population variance.
Compare the results in parts a) and c). Comment on which one you would choose if the data collection process is not a problem. What would be an even better approach to collect data in an attempt to answer the question of interest?
Method 1
Method 2
79
71
66
43
57
58
91
78
42
20
59
56
Explanation / Answer
(a)
Let the scores of method 1 and method 2 are stored in x and y vectors.
x = c(79,66,57,91,42,59)
y = c(71,43,58,78,20,56)
Use mean() and var() function in R to calculate the mean and standard deviation of scores of method 1 and method 2.
mean(x)
mean(y)
sqrt(var(x))
sqrt(var(y))
Mean of method 1 = 65.67
Mean of method 2 = 54.33
Standard deviation of method 1 = 17.32
Standard deviation of method 2 = 20.79
Null Hypothesis H0: The performance scores of teaching method 1 and method 2 are equal.
Alternative Hypothesis H0: The performance scores of teaching method 1 and method 2 are different.
The standard error (SE) of the sampling distribution.
SE = sqrt[ (s12/n1) + (s22/n2) ]
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
SE = sqrt[ (17.322/6) + (20.792/6) ] = 11.07832
The degrees of freedom (DF) is:
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (17.322/6 + 20.792/6)2 / { [ (17.322 / 6)2 / (6 - 1) ] + [ (20.792 / 6)2 / (6 - 1) ] }
= 9.668 = 10 (Rounded to nearest integer)
t = (Mean of method 1 - Mean of method 2) / SE
= (65.67 - 54.33)/ 11.07832 = 1.023621
P-value for t = 1.023621 and df = 10 is 0.1651
For two tail hypothesis, P - value = 2 * 0.1651 = 0.3302
As, the p-value is greater than significance level of 0.01, we fail to reject the null hypothesis and conclude that the performance scores of teaching method 1 and method 2 are equal.
T value at 99% confidence interval at DF = 10 is 3.17
Mean difference = 65.67 - 54.33 = 11.34
99% confidence interval is given as,
(11.34 - 3.17*11.07832, 11.34 + 3.17*11.07832)
= (-23.77827, 46.45827)
(b)
In R, we use t.test() funtion to confirm the results.
t.test(x,y,conf.level = 0.99)
Welch Two Sample t-test
data: x and y
t = 1.026, df = 9.6833, p-value = 0.3298
alternative hypothesis: true difference in means is not equal to 0
99 percent confidence interval:
-23.93371 46.60038
sample estimates:
mean of x mean of y
65.66667 54.33333
The results in R is consistent with part (a)
(c)
We will be conducting matched-pairs t-test.
The difference of scores of method 1 and method 2 = 8, 23, -1, 13, 22, 3
Mean difference = 11.34
Standard deviation of difference = 9.852
Null Hypothesis H0: The performance scores of teaching method 1 and method 2 are equal.
Alternative Hypothesis H0: The performance scores of teaching method 1 and method 2 are different.
The standard error (SE) of the sampling distribution = SD / sqrt(n) = 9.852/sqrt(6) = 4.022
Degree of freedom = n - 1 = 6 - 1 = 5
t = 11.34/4.022 = 2.819
P-value for t = 2.819 and DF = 5 is 0.0186
For two tail test, P-value = 0.0186 * 2 = 0.0372
As, p-value is greater than 0.01, we fail to reject the null hypothesis and conclude that the performance scores of teaching method 1 and method 2 are equal.
T value at 99% confidence interval at DF = 5 is 4.03
99% confidence interval is given as,
(11.34 - 4.03*4.022, 11.34 + 4.03*4.022)
= (-4.86866, 27.54866)
d)
In R, we use t.test() funtion with below option to confirm the results.
t.test(x,y,conf.level = 0.99,paired = TRUE)
Paired t-test
data: x and y
t = 2.8177, df = 5, p-value = 0.03721
alternative hypothesis: true difference in means is not equal to 0
99 percent confidence interval:
-4.884594 27.551261
sample estimates:
mean of the differences
11.33333
The result in R is consistent with the results in part (c)
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