A study was done on proctored and nonproctored tests. The results are shown in t

Question

A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple random in samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. Reject H_0. There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. Fail to reject H_0. There is not sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. Fail to reject H_0. There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. Construct a confidence interval suitable for testing the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.

Explanation / Answer

Solution:

alpha = 0.01
n1 = 35 , x1 = 75.55 , s1 = 10.99
n2 = 31 , x2 = 84.46 , s2 = 20.44

H0: 1 = 2
H1: 1 < 2

Standard error SE = (s1^2/n1+s2^2/n2)
= (10.99^2/35 + 20.44^2/31)
= 4.11437
test statistic t = (x1-x2) / SE
t = (75.55-84.46) / 4.11437
t = -2.165

degree of freedom = min(n1-1, n2-1) = 30
Since it is a 1-tailed test hence
P-value = 0.01925

Here P-value > 0.01
C) Fail to reject H0. There is not sufficient evidence to support the claim that students taking nonproctored tests get higher mean score than those taking proctored tests.

b) t(/2,df) = t(0.005, 30) = 2.749
confidence interval:
(xbar1-xbar2) ± [t(/2,df) * SE]
(75.55-84.46) ± [2.749 * 4.11437]
-8.91 ± 11.310
-20.22 <1-2< 2.4

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