SOLUTION: 1. To estimate with 95% confidence the average number of kilometres that students living off-campus commute to classes every day, a random sample of 20 students was taken and produ
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Question 1207888: 1. To estimate with 95% confidence the average number of kilometres that students living off-campus commute to classes every day, a random sample of 20 students was taken and produced a mean equal to 5.2 km and a standard deviation of 3.05 km. Under these circumstances, what is the appropriate critical value?
a. 2.093
b. 1.960
c. 2.086
d. 1.729 Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
n = 20 = sample size
xbar = 5.2 = sample mean
s = 3.05 = sample standard deviation
sigma = population standard deviation = unknown
We don't know the value of sigma and n > 30 is not the case, so we must use the T distribution.
df = degrees of freedom
df = n-1
df = 20-1
df = 19
You could use a stats calculator like a TI83 to find the T critical value.
However, I'll use a T table such as this https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
Such a table can be found at the back of your stats textbook.
Highlight the df = 19 row and the column that mentions "confidence level = 95%" (mentioned at the bottom of the table)
The intersection of this row and column yields the approximate t critical value t = 2.093
What does this tell us?
It tells us that P(-2.093 < t < 2.093) = 0.95 approximately when df = 19.
The 0.95 is the area of the main body while 1-0.95 = 0.05 is the combined area of the two tails.
The 0.95 refers to the confidence level 95%.
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