Respuesta :
Answer:
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]
