Question

The I-Recognition Test is a 100- question intellectual threshold test, designed to determine a bottom score of intellectual proficiency for all Federation crew-potential staff. All answers are graded as either correct or incorrect. They want to compare the two groups on I-Recognition Test scores to see if Group A has higher scores than Group B. One theory is that the exposure to certain rays in the nebulae may have caused an augmentation of intellectual functioning among the crew.

Group A: n = 30, x = 22

Group B: n = 30, x = 18

1. Determine whether the samples are independent. Explain why.

2. Determine whether the sample sizes are sufficiently large to conduct the hypothesis test.

3. Set up the null and alternative hypothesis statements:

4. List all computer/calculator steps used in your calculations.

5. Determine the conclusion: Is there sufficient statistical evidence to support the claim that Group A scoring is higher than Group B scoring?

List technology steps used to determine this. Use the p-value approach, given an alpha-level of .05.

6. Find the 95% confidence interval for the difference between the two sample proportions. Interpret the results in the context of the experiment. Be specific to the problem.

Answer 1

Given that,
sample one, x1 =22, n1 =30, p1= x1/n1=0.733
sample two, x2 =18, n2 =30, p2= x2/n2=0.6
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.733-0.6)/sqrt((0.667*0.333(1/30+1/30))
zo =1.095
| zo | =1.095
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =1.095 & | z α | =1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.0954 ) = 0.13666
hence value of p0.05 < 0.13666,here we do not reject Ho
ANSWERS
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i.
the samples are independent because they are random samples
ii.
yes,
whether the sample sizes are sufficiently large to conduct the hypothesis test.
iii.
null, Ho: p1 = p2
alternate, H1: p1 > p2
iv.
test statistic: 1.095
critical value: 1.645
decision: do not reject Ho
v.
p-value: 0.13666
we do not have enough evidence to support the claim that Group A scoring is higher than Group B scoring
vi.
TRADITIONAL METHOD
given that,
sample one, x1 =22, n1 =30, p1= x1/n1=0.733
sample two, x2 =18, n2 =30, p2= x2/n2=0.6
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.733*0.267/30) +(0.6 * 0.4/30))
=0.12
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.12
=0.236
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.733-0.6) ±0.236]
= [ -0.103 , 0.369]
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DIRECT METHOD
given that,
sample one, x1 =22, n1 =30, p1= x1/n1=0.733
sample two, x2 =18, n2 =30, p2= x2/n2=0.6
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.733-0.6) ± 1.96 * 0.12]
= [ -0.103 , 0.369 ]
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interpretations:
1) we are 95% sure that the interval [ -0.103 , 0.369] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean P1-P2