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]

Option a. [52.78,64.22]

Step-by-step 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;

                ~

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

P(-2.110 < < 2.110) = 0.95

P(-2.110 < < 2.110) = 0.95

P(-2.110 * < < 2.110 * ) = 0.95

P(X bar - 2.110 * < < X bar + 2.110 * ) = 0.95

95% confidence interval about = [ X bar - 2.110 * , X bar + 2.110 * ]

                                                   = [ 58.5 - 2.110 * , 58.5 + 2.110 * ]

                                                   = [ 52.78 , 64.22 ]

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

arithmetic sequence: a sequnce in which the diffrence between any two consecutive terms is the same

step-by-step explanation:

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step-by-step explanation:

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