Number of cases 20
To find the mean, add all of the observations and divide by 20
Mean 346.5
Squared deviations
(0-346.5)^2 = (-346.5)^2 = 120062.25
(260-346.5)^2 = (-86.5)^2 = 7482.25
(356-346.5)^2 = (9.5)^2 = 90.25
(403-346.5)^2 = (56.5)^2 = 3192.25
(536-346.5)^2 = (189.5)^2 = 35910.25
(0-346.5)^2 = (-346.5)^2 = 120062.25
(268-346.5)^2 = (-78.5)^2 = 6162.25
(369-346.5)^2 = (22.5)^2 = 506.25
(428-346.5)^2 = (81.5)^2 = 6642.25
(536-346.5)^2 = (189.5)^2 = 35910.25
(268-346.5)^2 = (-78.5)^2 = 6162.25
(396-346.5)^2 = (49.5)^2 = 2450.25
(469-346.5)^2 = (122.5)^2 = 15006.25
(536-346.5)^2 = (189.5)^2 = 35910.25
(162-346.5)^2 = (-184.5)^2 = 34040.25
(338-346.5)^2 = (-8.5)^2 = 72.25
(403-346.5)^2 = (56.5)^2 = 3192.25
(536-346.5)^2 = (189.5)^2 = 35910.25
(536-346.5)^2 = (189.5)^2 = 35910.25
(130-346.5)^2 = (-216.5)^2 = 46872.25
Add the squared deviations and divide by 19
Variance = 551547/19
Variance 29028.7895
Standard deviation = sqrt(variance) = 170.3784
a) 95 % limits
Since we do not know the population standard deviation and estimate it by the sample, we use the t-distribution with 20-1=19 degrees of freedom assuming normality.
Sample mean 346.5
Standard deviation = 170.3784
Standard error of mean = sd / sqrt(n)
SE = 170.3784/4.4721
Standard error of mean 38.0978
Confidence limits 346.5-(38.0978)(2.093)
and 346.5+(38.0978)(2.093)
95 % limits for mean is (266.7614, 426.2386)
b) Construct a histogram. Group the values into intervals and see if the data represent a bell-shaped curve. If it doesn't, normality would be an issue.
c) sample size n = (2.57 * 170.3784 / 10) ^2
n=1917.32 = 1917
d) The sample size is reasonable.
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