Question

A random sample of non-English majors at a selected college was used in a study to see if the student retained more from reading a 19th-century novel or by watching it in DVD form. Each student was assigned one novel to read and a different one to watch, and then they were given a 100-point written quiz on each novel. The test results are shown. Book 90 80 90 75 80 90 84 DVD 95 82 80 80 70 85 90 Use a 0.05 significance level to test the claim that the students score higher after watching the DVD.

1.Find the P-value.

2.Based on the P-value found , state your conclusion about the null hypothesis.

3.Based on your conclusion, state the conclusion about the claim.

4.Construct the 90% confidence interval to estimate the mean differences of the scores.

5.Based on the confidence interval, state the conclusion about the means.

Answer 1

Sample	Book	DVD	diff(d)=sample1-sample2	(d-dbar)^2
1	90	95	5	36
2	80	82	2	9
3	90	80	-10	81
4	75	80	5	36
5	80	70	-10	81
6	90	85	-5	16
7	84	90	6	49
		SUM	-7	308

t-Test: Paired Two Sample for Means
	Book	DVD
Mean	84.14286	83.14286
Variance	36.80952	64.14286
Observations	7	7
Pearson Correlation	0.51058
Hypothesized Mean Difference	0
df	6
t Stat	0.369274
P(T<=t) one-tail	0.362302
t Critical one-tail	1.439756
P(T<=t) two-tail	0.724605
t Critical two-tail	1.94318

Mean(d)=dbar=sum(d)/n   -1
standard deviation(sd)=sqrt(sum(d-dbar)^2/n-1)   7.16472842
n   7

test statistics=(dbar/(Sd/sqrt(n)) = -0.3693

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1)

P-value

p-value=2*p(t>|test statistics|) =

here df=n-1=6 using excel

2*p(t>|test statistics|) = 2*(1-T.DIST(0.3693,6,1)) =0.7245

P-value=0.7245

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2.Based on the P-value found , state your conclusion about the null hypothesis.

P-value=0.7245 >  0.05 we do not reject H0

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3

.Based on your conclusion, state the conclusion about the claim.

we do not reject H0 hence

There is no difference in reading a 19th-century novel or by watching it in DVD form

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4.Construct the 90% confidence interval to estimate the mean differences of the scores.

for 90 % confidence level with degree of freedom (n-1)=6  
\alpha=1-c%=1-0.90=0.1   0.1
degrees of freedom (n-1)   6
t=critical value at 90 % with (n-1) degree of freedom   1.94318
Margin of error =t*s/sqrt(n) =5.2622
Lower limit =dbar-ME = -1 - 5.2622 = -6.2622
Upper limit =dbar+ME = -1 + 5.2622 = 4.2622

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5.Based on the confidence interval, state the conclusion about the means.

If we take100 samples each of size 7 and calculate 90% confidence interval,.then about 90 out 100 confidence interval,will contain true difference.

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