. The digital music players produced by a large Canadian manufacturer during the
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Question
. The digital music players produced by a large Canadian manufacturer during the first two months of 2010 were of poor quality. A random sample of 100 players was obtained during this time and the players were tested. It was determined that 25 of them were defective. As a result, quality control standards were then tightened. A random sample of 100 players was taken during the next two months, 11 of which were defective. Is there sufficient evidence to indicate that the new quality control standards were effective in reducing the proportion of defective digital music players? To answer this question, conduct a hypothesis test at the 5% level of significance.5. The digital music players produced by a large Canadian manufacturer during the first two months of 2010 were of poor quality. A random sample of 100 players was obtained during this time and the players were tested. It was determined that 25 of them were defective. As a result, quality control standards were then tightened. A random sample of 100 players was taken during the next two months, 11 of which were defective. Is there sufficient evidence to indicate that the new quality control standards were effective in reducing the proportion of defective digital music players? To answer this question, conduct a hypothesis test at the 5% level of significance.
Explanation / Answer
p1= x1/n1= 25/100= 0.25
p2= x2/n2= 11/100= 0.11
alpha= 0.05
Null Hypothesis H0: P1= P2
Alternative Hypothesis Ha: P1 >P2 (right tailed test)
For sample 1, we have that the sample size is n1=100, the number of favorable cases is X1=25, so then the sample proportion is p^1=n11/X1=25/100=0.25
For sample 2, we have that the sample size is n2=100, the number of favorable cases is X2=11, so then the sample proportion is p^2=n2/X2=11/100=0.11
The value of the pooled proportion is computed as p¯=n1+n2/X1+X2=25+11/200=0.18
Also, the given significance level is =0.05.
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho:p1=p2
Ha:p1>p2
This corresponds to a right-tailed test, for which a z-test for two population proportions needs to be conducted.
(2) Rejection Region
Based on the information provided, the significance level is lpha = 0.05=0.05, and the critical value for a right-tailed test is zc=1.64.
The rejection region for this right-tailed test is R={z:z>1.64}
(3) Test Statistics
The z-statistic is computed as follows:
z=p¯(1p¯)(1/n1+1/n2)p^1p^2=0.18(10.18)(1/100+1/100)0.250.11=2.577
(4) Decision about the null hypothesis
Since it is observed that 2.577>zc=1.64, it is then concluded that the null hypothesis is rejected.
Using the P-value approach: The p-value is p=0.005, and since p=0.005<0.05, it is concluded that the null hypothesis is rejected.
(5) Conclusion
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population proportion p1 is greater than p2, at the 0.05 significance level.
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