Researchers studied the friction that occurs in the paper-feeding process of a p
ID: 3238319 • Letter: R
Question
Researchers studied the friction that occurs in the paper-feeding process of a photocopier In a stack of 100 sheets of paper, the feeding process was successful 94 times. The success rate of the feeder is designed to be .90. Test to determine if the success rate of the feeder exceeds .90 at alpha = 0.05. Test statistic Z =p hat - p_0/squareroot p_0 middot (1-p_0)/n and Confidence Interval P hat plusminus z_alpha/2 squareroot p hat q hat/n State the null and alternative hypothesis. State your conclusion for your hypothesis test. Give the 95% confidence Interval. Give a conclusion for the confidence interval test.Explanation / Answer
Answer to part a)
Hypothesis:
.
Answer to part b)
Given:
.
Standard error SE = sqrt(P *(1-P)/n)
SE = sqrt( 0.90 *0.10 /100)
SE = 0.03
.
On pluggging the values in the formula of Test Statistic we get:
Z = (0.94 - 0.90) / 0.03
Z = 1.33
.
The Pvalue can be found with the help of the Z table
the area under the right tail is the P value in this case because the test is a right tailed test
Thus P value = 0.0918
.
Inference:
Given : alpha = 0.05
Since P value 0.0918 > alpha 0.05 , we fail to reject the null hypothesis
.
Conclusion: Thus we conclude that the success rate is equal to 0.90 . The claim that it exceeds 0.90 is not supported by the sample data. Hence we do not accept the claim.
.
Answer to part c)
P^ - Z * SE , P^ + Z * SE
Value of Z critical for 95% confidence level is 1.96
.
On pluggging the values we get:
0.94 - 1.96 * 0.03 , 0.94 + 1.96 *0.03
0.94 - 0.0588 , 0.94 + 0.0588
0.8812 , 0.9988
.
Answer to part d)
This confidence interval includes 0.90
0.8812 < 0.90 < 0.9988
Thus it supports the null , and hence we fail to reject the null
Thus there is no significant evidence to supoor the claim that the success rate exceeds 0.90
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