Peter, a eacher in a college, randomly his class to complete an assigned task, the sample mean completion time is found. Using this x, Peter found that the 95% confidence interval for the true average time in completing the task is (17.278, 28.722) minutes. Peter thinks that this interval is too wide, that means the speed of students doing an in-class exercise has to be improved. Therefore, he thinks a confidence interval of total width 5 minutes will be more desirable. (Hints: width of C.I. = upper limit minus lower limit) (a) Using the same value of x, find the new confidence interval in this case. (b) Find the confidence level for the interval in part (a).

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Peter, a statistics teacher in a community college, randomly selected thirty-five students in his class
to complete an assigned task, the sample mean completion time is found. Using this x, Peter found
that the 95% confidence interval for the true average time in completing the task is (17.278, 28.722)
minutes. Peter thinks that this interval is too wide, that means the speed of students doing an in-class
exercise has to be improved. Therefore, he thinks a confidence interval of total width 5 minutes will
be more desirable.
(Hints: width of C.I. = upper limit minus lower limit)
(a) Using the same value of x, find the new confidence interval in this case.
(b) Find the confidence level for the interval in part (a).
Transcribed Image Text:Peter, a statistics teacher in a community college, randomly selected thirty-five students in his class to complete an assigned task, the sample mean completion time is found. Using this x, Peter found that the 95% confidence interval for the true average time in completing the task is (17.278, 28.722) minutes. Peter thinks that this interval is too wide, that means the speed of students doing an in-class exercise has to be improved. Therefore, he thinks a confidence interval of total width 5 minutes will be more desirable. (Hints: width of C.I. = upper limit minus lower limit) (a) Using the same value of x, find the new confidence interval in this case. (b) Find the confidence level for the interval in part (a).
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i found the z value of 0.8563 from standard normal table is (0.22+0.21)/2=0.215. So the p-value is 2x(0.215)=0.43. The confidence level for new interval is 57%. Why the answers is different to 61%.

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