A study was done to compare the effectiveness of distance learning with traditional classroom instruction. Fourteen students took a business administration course online, while 17 students took it in a classroom. The final exam scores were as follows. It is reasonable to assume that the samples come from populations that are approximately normal. Online
65 74 84 63 87 76 73
90 71 68 76 82 61 69

Classroom
70 79 55 74 82 70 80 73 77
54 82 80 81 68 58 69 92

Required:
a. Construct a 95% confidence interval for the difference between the mean scores for the two types of instruction.
b. An educator claims that both methods of instruction are equally effective. Does the confidence interval contradict this claim?

See below for answers and explanations

Step-by-step explanation:

Part A:

Check conditions for a 2-sample t-interval:

Random sample? √

Is the sample size no more than 10% of the population? √

Are both samples approximately normal? √

Construct the 95% confidence interval:

On a TI-84, click on "Stat", scroll to "Tests", and select "2-SampTInt". Put your data into 2 lists and set C-Level to be 0.95 to represent a 95% confidence level. Make sure the data isn't pooled, and press "Calculate" to obtain your results. You will see that the 95% confidence interval for the difference between the mean scores for the two types of instruction is {-6.042,8.1175}

Part B:

The confidence interval does not contradict the claim because 0 is contained within the 95% confidence interval, so it is quite possible that there is no difference between the mean scores for the two types of instruction.

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

hope it