SOLUTION: Statistics calculated from a sample of 22 observations are:
{screenshot for better representation of equation: http://imgur.com/a/Qc4VK }
∑ (n=22 and i=1) x = 1451
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-> SOLUTION: Statistics calculated from a sample of 22 observations are:
{screenshot for better representation of equation: http://imgur.com/a/Qc4VK }
∑ (n=22 and i=1) x = 1451
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Question 1084646: Statistics calculated from a sample of 22 observations are:
{screenshot for better representation of equation: http://imgur.com/a/Qc4VK }
∑ (n=22 and i=1) x = 1451
∑ (n=22 and i=1) x^2 = 106639
(a) Find the sample mean: I found this to be 1451/22
(b) What is the sample standard deviation?
(c) Assume that the population distribution is normal. Find a 95% confidence interval for the population mean. ( , ) Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website! Part (a)
Correct. The fraction 1451/22 approximates to 65.9545454545454
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Part (b)
Notation notes:
When I say "sigma(X^2)" I mean which in this case is 106639
Similarly, "sigma(X)" means which in this case is 1451
Using those values and n = 22, we can use the formula below to get
s = sqrt( (n*sigma(X^2)-(sigma(X))^2)/(n*(n-1)) )
s = sqrt( (22*106639-(1451)^2)/(22*(22-1)) )
s = sqrt( (22*106639-2105401)/(22*21) )
s = sqrt( (2346058-2105401)/(462) )
s = sqrt( (240657)/(462) )
s = sqrt( 520.902597402597 )
s = 22.8232906786598
which is the approximate answer to part (b)
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Part (c)
We're given n = 22 as the sample size.
Because n > 30 is not true and we don't know sigma (population standard deviation) we must use the T distribution.
The degrees of freedom (df) are
df = n-1
df = 22-1
df = 21
The value of xbar is approximately 65.9545454545454 as done in part (a)
Use a table such as this one to locate the df = 21 row.
I have marked this row with a red rectangle (see below)
In this df = 21 row, mark the value that is directly over top the "95%" confidence level
I marked this entire column with a blue rectangle (see below)
The red and blue rectangles intersect at the value 2.080
So the t critical value is t = 2.080
Earlier in part (b) we calculated the sample standard deviation to be approximately s = 22.8232906786598
We take these values (xbar = 65.9545454545454, t = 2.080, s = 22.8232906786598, n = 22) to plug into the formulas below
L = xbar - t*(s/sqrt(n))
L = 65.9545454545454 - 2.08*(22.8232906786598/sqrt(22))
L = 65.9545454545454 - 2.08*(22.8232906786598/4.69041575982343)
L = 65.9545454545454 - 2.08*(4.8659419222826)
L = 65.9545454545454 - 10.1211591983478
L = 55.8333862561976
U = xbar + t*(s/sqrt(n))
U = 65.9545454545454 + 2.08*(22.8232906786598/sqrt(22))
U = 65.9545454545454 + 2.08*(22.8232906786598/4.69041575982343)
U = 65.9545454545454 + 2.08*(4.8659419222826)
U = 65.9545454545454 + 10.1211591983478
U = 76.0757046528932
The 95% confidence interval for the mean (mu) is (L,U) = (55.8333862561976,76.0757046528932), which is an approximate interval.
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