Respuesta :
Answer:
(11.76, 12.84)
Step-by-step explanation:
Given that we can assume that the distribution of individual student enrollment units at this college is approximately normal.
Sample mean =12.3
sample std dev s = 1.9 units
Sample size = 47
Std error = [tex]\frac{1.9}{\sqrt{47} } \\=0.277[/tex]
Z critical value for 95% = 1.96
Margin of error = [tex]1.96*0.277=0.543[/tex]
Confidence interval =
[tex](12.3-0.543, 12.3+0.543)\\= (11.757, 12.843)[/tex]
=(11.76, 12.84)
Using the t-distribution, it is found that the 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.9).
In this question, we are given the standard deviation for the sample, thus, the t-distribution is used to build the confidence interval.
The information for the sample are as follows:
- 47 students, thus [tex]n = 47[/tex]
- Mean of 12.2, thus [tex]\overline{x} = 12.3[/tex].
- Standard deviation of 1.6, thus [tex]\sigma = 1.9[/tex].
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
First, we find the number of degrees of freedom, which is one less than the sample size, thus df = 46.
Then, we find the critical value for a 95% confidence interval, with 46 df, which is t = 2.0129.
Then, the interval is given by:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 12.3 - 2.0129\frac{1.9}{\sqrt{47}} = 11.7[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 12.3 + 2.0129\frac{1.9}{\sqrt{47}} = 12.9[/tex]
The 95% confidence interval for the average number of units that students in their college are enrolled in is (11.7, 12.9).
A similar problem is given at https://brainly.com/question/13769784
