Homework Sourse1 At the 05 level test the claim that the proportion of defe
1. At the .05 level, test the claim that the proportion of defects for CD burners is 0.3. A sample of 200 randomly selected CD burners had 57 defects. List the null and alternate hypotheses.
Find the critical values
find the test statistic
what is your conclusion?
2. A supermarket manager claims that the standard deviation for errors when a scanner is used is less then 200. At the 0.01 level test the claim if a sample of 20 tickets found a standard deviation of 189.
Select the null and alternate hypotheses
Find the critical value(s)
Find the test statistic
What is your conclusion?
3. In a clinical study of allergy drug, 115 of the 204 subjects reported experiencing significant relief from their symptoms. At the 0.01 significance level, test the claim that more than half of all those using the drug experience relief.
List the null and alternate hypotheses.
Find the critical value(s)
Find the test statistic.
What is your conclusion?
Solution
(1)List the null and alternate hypotheses.
Null hypothesis: p=0.3
Alternative hypothesis: p not equal to 0.3
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Find the critical values
It is a two-tailed test.
Given a=0.05, the critical values are Z(0.025) = -1.96 or 1.96 (from standard normal table)
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find the test statistic
Z=(phat-p)/sqrt(p*(1-p)/n)
=(57/200-0.3)/sqrt(0.3*0.7/200)
=-0.46
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what is your conclusion?
Since Z=-0.46 is between -1.96 and 1.96, we do not reject the null hypothesis.
So we can conclude that the proportion of defects for CD burners is 0.3.
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(2)Select the null and alternate hypotheses
Let o^2 be the variance
Null hypothesis: o^2=200
Alternative hypothesis: o2^2<200
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Find the critical value(s)
It is a left-tailed test.
The degree of freedom =n-1=20-1=19
Given a=0.01, the critical value of chisquare with 0.01 and df=19 is 7.63 (from chisquare table)
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Find the test statistic
chisquare = (n-1)*s^2/o^2
=(20-1)*189^2/200^2
=16.97
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What is your conclusion?
Since 16.97 is larger than 7.63, we do not reject the null hypothesis.
So we can not conclude that the standard deviation for errors when a scanner is used is less then 200.
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(3)List the null and alternate hypotheses.
Null hypothesis: p=0.5
Alternative hypothesis: p>0.5
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Find the critical value(s)
It is a right-tailed test.
Given a=0.01, the critical value is Z(0.01) =2.33 (from standard normal table)
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Find the test statistic.
Z=(phat-p)/sqrt(p*(1-p)/n)
=(115/204-0.5)/sqrt(0.5*0.5/204)
=1.82
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What is your conclusion?
Since Z=1.82 is less than 2.33, we do not reject the null hypothesis.
So we can not conclude that more than half of all those using the drug experience relief.
