Twenty university students chosen at random from students enrolled at Carthage College took part in an experiment testing the effect of watching ABC News versus CNN on knowledge of world events. The 20 participants were divided randomly into two equal groups of 10 students each. One group was assigned to watch ABC and the other was assigned to watch CNN. Participants then watched their assigned station (ABC or CNN) for 3 hours a night for 4 weeks. At the end of the 4 week period they were administered a test of world knowledge. The results were as follows:XbarABC = 104; sABC = 12XbarCNN = 99; sCNN = 10Compute the 95% confidence interval around the difference between the two groups.

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Asked Sep 25, 2019
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  • Twenty university students chosen at random from students enrolled at Carthage College took part in an experiment testing the effect of watching ABC News versus CNN on knowledge of world events. The 20 participants were divided randomly into two equal groups of 10 students each. One group was assigned to watch ABC and the other was assigned to watch CNN. Participants then watched their assigned station (ABC or CNN) for 3 hours a night for 4 weeks. At the end of the 4 week period they were administered a test of world knowledge. The results were as follows:

XbarABC = 104; sABC = 12

XbarCNN = 99; sCNN = 10

  1. Compute the 95% confidence interval around the difference between the two groups.
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Expert Answer

Step 1

We need to construct the 95% confidence interval for the difference between the population means

μ1 - μ2 , for the case that the population standard deviations are not known and sample size is <30. The following information has been provided about each of the samples

Step 2

Assuming sample 1 =ABC

Sample 2 = CNN

 

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nl 10 n2 10 xl 104 x2 99 s1 12 s2 10

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Step 3

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are 

 df=n1​+n2​−2=10+10−2=18

Given confidence level = CL = 0.95, so α= 1 – CL = 1-0.95 = 0.05

The critical value for α=0.05 and df = 18 is tc

...
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4a2n-1=2.101 ... Using t-table

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