Question

Question 21 [t-interval] a random sample of size 18 is drawn from a population that is normally distributed. the sample mean is 58.5, and the sample standard deviation is found to be 11.5. determine a 95% confidence interval about population mean.

a. [52.78,64.22]
b. [53.78,63.22]
c. [53.18,63.81]
d. [54.04,62.96]

Answer 1

Option a. [52.78,64.22]

Explanation:

We are given that a random sample of size 18 is drawn from a population that is normally distributed.

Sample mean is 58.5, and the sample standard deviation is found to be 11.5 i.e., X bar = 58.5 and s = 11.5

The pivotal quantity for calculating 95% confidence interval is;

`(Xbar - \mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

So, 95% confidence interval about population mean is given by;

P(-2.110 <
`t_1_7` < 2.110) = 0.95

P(-2.110 <
`(Xbar - \mu)/((s)/(√(n) ) )` < 2.110) = 0.95

P(-2.110 *
`(s)/(√(n) )` <
`{Xbar - \mu}` < 2.110 *
`(s)/(√(n) )` ) = 0.95

P(X bar - 2.110 *
`(s)/(√(n) )` <
`\mu` < X bar + 2.110 *
`(s)/(√(n) )` ) = 0.95

95% confidence interval about
`\mu` = [ X bar - 2.110 *
`(s)/(√(n) )` , X bar + 2.110 *
`(s)/(√(n) )` ]

= [ 58.5 - 2.110 *
`(11.5)/(√(18) )` , 58.5 + 2.110 *
`(11.5)/(√(18) )` ]

= [ 52.78 , 64.22 ]

Therefore, 95% confidence interval about population mean is [52.78 , 64.22].