Answer:
a) The margin of error for a 90% confidence interval when n = 14 is 18.93.
b) The margin of error for a 90% confidence interval when n=28 is 12.88.
c) The margin of error for a 90% confidence interval when n = 45 is 10.02.
Step-by-step explanation:
The t-distribution is used to solve this question:
a) n = 14
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 = 14 - 1 = 13
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of
. So we have T = 1.7709
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The margin of error for a 90% confidence interval when n = 14 is 18.93.
b) n = 28
27 df, T = 1.7033

The margin of error for a 90% confidence interval when n=28 is 12.88.
c) The margin of error for a 90% confidence interval when n = 45 is
44 df, T = 1.6802

The margin of error for a 90% confidence interval when n = 45 is 10.02.
Answer:
a) 28
b)
Class Limits Class Boundaries Midpoint f Rf Cf
10 - 37 9.5-37.5 23.5 7 0.096 7
38 - 65 37.5-65.5 51.5 25 0.342 32
66 - 93 65.5-93.5 79.5 26 0.356 58
94 - 121 93.5-121.5 107.5 9 0.123 67
122 - 149 121.5-149.5 135.5 5 0.068 72
150 - 177 149.5-177.5 163.5 0 0.0 72
178 - 205 177.5-20.5 191.5 1 0.014 73
Total 73 1.0
c)
The histogram is in attached file.
Step-by-step explanation:
a)
The width of class interval=range/desired number of classes
The width of class interval=(maximum-minimum)/7
The width of class interval=(200-10)/7
The width of class interval=190/7
The width of class interval=27.14=28 (rounded to next whole number)
Part(b) and part (c) is explained in attached word document.
You can expect 76 students in 2018 because from 2014 to 2015 it went up 6, then 8, then 10 so it would increase 12 students from 2017 to 2018
no -17 is greater than -6