A simple random sample of size n is drawn. The sample mean. x. is found to be 18

Question

A simple random sample of size n is drawn. The sample mean. x. is found to be 18.8, and the sample standard deviation, s. is found to be 4.1. Construct a 95% confidence interval about p if the sample size, n, is 34. Lower bound:: Upper bound: Construct a 95% confidence interval about p if the sample size, n, is 51. Lower bound:: Upper bound: (Use ascending order. Round to two decimal places as needed.) How does increasing the sample size affect the margin of error, E The margin of error does not change. The margin of error decreases. The margin of error increases. Construct a 99% confidence interval about p if the sample size, n, is 34. Lower bound:: Upper bound: (Use ascending order. Round to two decimal places as needed.) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the size of the margin of error. E The margin of error increases. The margin of error does not change. The margin of error decreases. If the sample size is 18. what conditions must be satisfied to compute the confidence interval The sample data must come from a population that is normally distributed with no outliers. The sample size must be large and the sample should not have any outliers. The sample must come from a population that is normally distributed and the sample size must be large.

Explanation / Answer

a.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=18.8
Standard deviation( sd )=4.1
Sample Size(n)=34
Confidence Interval = [ 18.8 ± t a/2 ( 4.1/ Sqrt ( 34) ) ]
= [ 18.8 - 2.035 * (0.703) , 18.8 + 2.035 * (0.703) ]
= [ 17.37,20.23 ]

b.
Mean(x)=18.8
Standard deviation( sd )=4.1
Sample Size(n)=51
Confidence Interval = [ 18.8 ± t a/2 ( 4.1/ Sqrt ( 51) ) ]
= [ 18.8 - 2.009 * (0.57) , 18.8 + 2.009 * (0.57) ]
= [ 17.65,19.95 ]

c.
Margin of erroe will decrease

d.

Standard deviation( sd )=4.1
Sample Size(n)=34
Confidence Interval = [ 18.8 ± t a/2 ( 4.1/ Sqrt ( 34) ) ]
= [ 18.8 - 2.733 * (0.7) , 18.8 + 2.733 * (0.7) ]
= [ 16.88,20.72 ]

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