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 Determine whether the samples are independent. Explain why. Determine whether the sample sizes are sufficiently large to conduct the hypothesis test. Set up the null and alternative hypothesis statements: List all computer/calculator steps used in your calculations. 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. 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

Answer:-

Given that:-

Given that,
sample one, =22, =30, ==0.733
sample two,
null, Ho: =
alternate,
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 > 1.645
we use test statistic (z) =



critical value
the value of |z α| at lo s 0.05% is 1.645
we got =1.095 & | z α | =1.645
make decision
hence value of and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.0954 ) = 0.13666
hence value of p 0.05 < 0.13666,here we do not reject Ho

the samples are independent because they are random samples

yes,
whether the sample sizes are sufficiently large to conduct the hypothesis test.

null, Ho:
alternate,

test statistic: 1.095
critical value: 1.645
decision: do not reject Ho

p-value: 0.13666
we do not have enough evidence to support the claim that Group A scoring is higher than Group B scoring

TRADITIONAL METHOD
given that,
sample one,
sample two,
I.
standard error =
where
= proportion of both sample observation
= sample size
standard error =
II.
margin of error = Z a/2 * (standard error)
where,
= 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 = ± 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,
sample two,
CI =
where,
= proportion of both sample observation
= size of both group
a = 1 - (confidence Level/100)
= 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
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