100y−39,000=55(x−800)
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Answer:
The 95% confidence interval for the average number of units that students in their college are enrolled in is between 11.7 and 12.5 units.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 45 - 1 = 44
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of . So we have T = 2
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 12.1 - 0.4 = 11.7 units
The upper end of the interval is the sample mean added to M. So it is 12.1 + 0.4 = 12.5 units
The 95% confidence interval for the average number of units that students in their college are enrolled in is between 11.7 and 12.5 units.