Use the given sample data to construct the indicated confidence interval for the population mean. The principal randomly selected six students to take an aptitude test. Their scores were: 71.6 81.0 88.9 80.4 78.1 72.0 Determine a 90% confidence interval for the mean score for all students. Group of answer choices

The 90% confidence interval

(74.71, 82.63)

Step-by-step explanation:

Confidence Interval Formula is given as:

Confidence Interval = μ ± z (σ/√n)

Where

μ = mean score

z = z score

N = number of the population

σ = standard deviation

The mean is calculated as = The average of their scores

N = 6 students

(71.6 + 81.0 + 88.9 + 80.4 + 78.1 + 72.0 )/ 6

Mean score = 472/6

= 78.666666667

≈ 78.67

We are given a confidence interval of 90% therefore the

z score = 1.645

Standard Deviation for the scores =

s=(x -σ)²/ n - 1 =(71.6 - 78.67)²+(81.0 - 78.67)²+(88.9 - 78.67)² + (80.4 - 78.67)²+ (78.1 - 78.67)²+( 72.0 - 78.67)2/ 6 - 1

= 5.886047531

= 5.89

The confidence interval is calculated as

= μ ± z (σ/√N)

= 78.67 ± 1.645(5.89/√6)

= 78.67 ± 3.9555380987

The 90% confidence interval

is :

78.67 + 3.9555380987 = 82.625538099

78.67 - 3.9555380987 = 74.714619013

Therefore, the confidence interval is approximately between

(74.71, 82.63)

is there answer choices?