The principal randomly selected six students to take an aptitude test. Their scores were: 76.5 85.2 77.9 83.6 71.9 88.6 Determine a 90% confidence interval for the mean score for all students.

90% Confidence Interval = [75.50, 85.74]

Step-by-step explanation:

The principal randomly selected six students to take an aptitude test. Their scores were: 76.5 85.2 77.9 83.6 71.9 88.6 Determine a 90% confidence interval for the mean score for all students.

Step 1

We find the Mean and Standard deviation

Mean = Sum of terms/Number of terms

Mean = 76.5 + 85.2 + 77.9 + 83.6 + 71.9 + 88.6/6

Mean = 483.7/6

Mean = 80.61666667

Approximately = 80.62

Stand Deviation =

√(x - mean)²/n - 1

= √193.9483333/6 - 1

= √38.78966667

= 6.228135087

Approximately = 6.23

Step 2

If you look at the question, our number of samples is 6. This is a small sample size and it is less than 30 hence, the formula for Confidence Interval that we would be using is

Confidence Interval = Mean ± t × Standard deviation/√n

Where t = test score of the 90% confidence interval

Degrees of freedom = n - 1

= 6 - 1 = 5

t score for a 90% confidence interval = 2.015

Hence, Confidence Interval =

80.62 ± 2.015 × 6.23/√6

80.62 ± 2.015 × 2.5433868496

80.62 ± 5.1249245019

Confidence interval

= 80.62 - 5.1249245019

= 75.495075498

≈ 75.50

= 80.62 + 5.1249245019

= 85.744924502

≈ 85.74

90% Confidence Interval = [75.50, 85.74]