Question

In: Statistics and Probability

A simple random sample of size n is drawn. The sample​ mean, x overbar​, is found...

A simple random sample of size n is drawn. The sample​ mean, x overbar​, is found to be 18.9​, and the sample standard​ deviation, s, is found to be 4.4.

a) Construct a 95​% confidence interval about mu if the sample​ size, n, is 34. Lower​ bound: nothing​; Upper​ bound: nothing ​(Use ascending order. Round to two decimal places as​ needed.)

​(b) Construct a 95​% confidence interval about mu if the sample​ size, n, is 61. Lower​ bound: nothing​; Upper​ bound: nothing ​(Use ascending order. Round to two decimal places as​ needed.) How does increasing the sample size affect the margin of​ error, E? A. The margin of error does not change. B. The margin of error increases. C. The margin of error decreases. ​

(c) Construct a 99​% confidence interval about mu if the sample​ size, n, is 34. Lower​ bound: nothing​; Upper​ bound: nothing ​(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? A. The margin of error decreases. B. The margin of error does not change. C. The margin of error increases. ​

(d) If the sample size is 14​, what conditions must be satisfied to compute the confidence​ interval? A. The sample size must be large and the sample should not have any outliers. B. The sample data must come from a population that is normally distributed with no outliers. C. The sample must come from a population that is normally distributed and the sample size must be large

Solutions

Expert Solution

We can see that 'n' changes and the (1 -)% confidence level changes. Rest is constant

= 18.9 Sx = 4.4

(1- )% is the confidence interval for population mean

Where Margin of error = C.v. * SE

=

0

a) Construct a 95​% confidence interval about mu if the sample​ size, n, is 34. Lower​ bound: nothing​; Upper​ bound: nothing ​(Use ascending order. Round to two decimal places as​ needed.)

Alpha =1 - 0.95 = 0.05

Therefore the C.V. =

=

= 2.0345 .............found using t-dist tables

Substituting the values we have

()

​(b) Construct a 95​% confidence interval about mu if the sample​ size, n, is 61. Lower​ bound: nothing​; Upper​ bound: nothing ​

Alpha =1 - 0.95 = 0.05

Therefore the C.V. =

=

= 2.0003 .............found using t-dist tables

Substituting the values we have

(Use ascending order. Round to two decimal places as​ needed.) How does increasing the sample size affect the margin of​ error, E? A. The margin of error does not change. B. The margin of error increases. C. The margin of error decreases. ​

We can easily see that the CI is narrower when n= 61. Increase 'n' increases accuracy so error reduces in return.

Can be mathematically tested as well.

(c) Construct a 99​% confidence interval about mu if the sample​ size, n, is 34. Lower​ bound: nothing​; Upper​ bound: nothing ​(Use ascending order. Round to two decimal places as​ needed.)

Alpha =1 - 0.99 = 0.01

Therefore the C.V. =

=

=2.7333 .............found using t-dist tables

Substituting the values we have

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? A. The margin of error decreases. B. The margin of error does not change. C. The margin of error increases. ​

Increasing the level of significance increases the strictness of test. So to get more accuracy it tries to get more possible values which widens the CI because MOE increases.

when we compare 'n' and level of confidence, they work in opposite direction. 'n' are the factual values so larger 'n' ensures estimation becomes closer to real value thus error reduces. level of confidence is the probability so increasing the probability means to incease the interval which enhances the chances of the true parameter being within the interval.

(d) If the sample size is 14​, what conditions must be satisfied to compute the confidence​ interval? A. The sample size must be large and the sample should not have any outliers. B. The sample data must come from a population that is normally distributed with no outliers. C. The sample must come from a population that is normally distributed and the sample size must be large

For n to be large it needs to be n > 20 or at least symmetrical.. So A can't be true. If the data is normal it need not have large sample size. So C is not true.


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