In: Statistics and Probability
Students in a statistics class are conducting a survey to estimate the mean number of units students at their college are enrolled in. The students collect a random sample of 46 students. The mean of the sample is 12.5 units. The sample has a standard deviation of 1.8 units.
What is the 95% confidence interval for the average number of units that students in their college are enrolled in? Assume that the distribution of individual student enrollment units at this college is approximately normal.
( , )
Your answer should be rounded to 2 decimal places.
Solution :
Given that,
= 12.5
s = 1.8
n = 46
Degrees of freedom = df = n - 1 = 46 - 1 = 45
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,45 = 2.014
Margin of error = E = t/2,df * (s /n)
= 2.014 * (1.8 / 46)
= 0.53
The 95% confidence interval estimate of the population mean is,
- E < < + E
12.5 - 0.53 < < 12.5 + 0.53
11.97 < < 13.03
(11.97 , 13.03)