Question

Ten randomly selected people took an IQ test A. The next day they took a very similar IQ test B. Their scores are shown in the table below.

Person	1	2	3	4	5	6	7	8	9	10
Test A	106	99	95	92	87	127	115	100	77	108
Test B	112	101	99	96	90	130	112	99	80	112

1. Use a 0.05 significance level to test the claim that, on average, people do better on the second test than they do on the first.

(a) State the null and alternative hypotheses. (Type muD for the mean of the differences in test A - test B.)
H0 :
H1 :

(b) Find the rejection region.
Reject H0 if t<

(c) Find the test statistic. (If necessary, round the sample mean and sample standard deviation to 2 decimal places. Also round your final answer to 2 decimal places.)
t =

(d) Do these samples provide significant evidence that, on average, people do better on the second test than they do on the first? (Type: Yes or No )



2. Construct a 95% confidence interval for the mean of the differences in test A - test B. (Round your final answer to 2 decimal places.)
(  ,  )

Answer 1

(a)

= mean of the differences in test A - test B

claim on average, people do better on the second test than they do on the first. : Test B > Test A i.e Test A -Test B < 0

< 0

Ho : = 0

Ha : < 0

Left tail test

(b) Find the rejection region.

Number of persons = Sample size : n=10

Degrees of freedom : n-1 = 10-1 =9

Significance level = 0.05

Critical value of left tailed test : -t0.05

For 9 degrees of freedom t0.05 = 1.833

Critical value of left tailed test : -t0.05 = -1.833
Reject H0 if t < -1.860

c) Find the test statistic. (If necessary, round the sample mean and sample standard deviation to 2 decimal places. Also round your final answer to 2 decimal places.)

di : Sample differnce in test score of student i (Test A - Test B)

Sample mean difference :

Sample standard deviation of difference

Person	Test A	Test B	d	(d-)	(d-)2
1	106	112	-6	-3.5	12.25
2	99	101	-2	0.5	0.25
3	95	99	-4	-1.5	2.25
4	92	96	-4	-1.5	2.25
5	87	90	-3	-0.5	0.25
6	127	130	-3	-0.5	0.25
7	115	112	3	5.5	30.25
8	100	99	1	3.5	12.25
9	77	80	-3	-0.5	0.25
10	108	112	-4	-1.5	2.25
		Total	-25		62.5
		Mean:	-25/10-=-2.5		(d-)2 = 62.5

Test Statistic t = -3.00

(d) Do these samples provide significant evidence that, on average, people do better on the second test than they do on the first? (Type: Yes or No )

Answer : Yes

As Calculated Value of t is less than Critical Value i.e. ( -3.0<-1.833 ); Reject Null Hypothesis. There is sufficient evidence to conclude that the difference < 0 i.e people do better on the second test than they do on the first

2. Construct a 95% confidence interval for the mean of the differences in test A - test B. (Round your final answer to 2 decimal places.)
(  ,  )

Formula for confidence interval for the mean of the differences

for 95% confidence interval = (100-95)/100 =0.05

/2 = 0.05/2 =0.025

t/2,n-1 = t0.025,9 = 2.2622

95% confidence interval for the mean of the differences in test A - test B.

95% confidence interval for the mean of the differences in test A - test B (-4.39,-0.62)

Answer

(-4.39,-0.62)