Two teaching methods (\"standard\" and \"new\") are compared based on a resu lt
ID: 3154114 • Letter: T
Question
Two teaching methods ("standard" and "new") are compared based on a resu lt of a reading the end a learning penod.Of a random sample of students, 10 are taught by new method by the standard method. The test results are given in the table. Use the table data to estimate the true mean difference between the test scores for method and the standard method. Use a 95% confidence interval. Interpret the result from part A. What assumptions must hold in order for the estimate to be valid Are they statisfiedExplanation / Answer
A. Estimate of true mean difference can be found out using confidence interval
Since, population size is less than 30, t statistic is calculated to estimate mean difference,
n1 = 10
n2 = 12
x1 (mean of 1st sample) = 76.4
x2 (mean of 2nd sample) = 72.33
s1 (sample standard deviation of 1st sample) = 5.835
s2 (sample standard deviation of 2nd sample) = 6.344
degree of freedom = 10+12-2 = 20
t(0.025,20) = 2.086
sp (pooled standard deviantion) = sqrt(((n1-1)s1^2+(n2-1)s2^2)/(n1+n2-2))
sp = 6.12
Confidence Interval = (x1-x2 - t(0.025,20)*sp*sqrt(1/n1+1/n2), x1-x2 + t(0.025,20)*sp*sqrt(1/n1+1/n2))
Substitute values in the above equation to get confidance interval,
CI = (-1.4, 9.53)
B. Interpretation of above result is that, we do not know the actual difference between population means, but using sample mean and standard deviation we can say that, 95% of the time differerence betwwen population mean would lie on the calculated interval
C. Assumptions are:
1) The two populations have the same variance.
2) The populations are normally distributed
3) Each value is sampled independently from each other value.
1) and 2) is not mentioned in the problem. So they are not satisfied.
3) is satisfied because random sample of students are used in the analysis
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