Question 1179328: At a certain university, the average attendance at basketball games has been 3025. Due to the dismal showing of the team this year, the attendance for the first 14 games has averaged only 2815 with a standard deviation of 485. The athletic director claims that the attendance is the same as last year. What is the test value needed to evaluate the claim?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! population mean assumed to be 3025
sample mean = 2815
standard deviation of sample = 485.
sample size = 14
t-score is indicated because standard deviation taken from sample rather than from population.
t-score = (x - m) / s
x is the raw score
m is the raw mean of the population
s = standard deviation of sample / sqrt(sample size) = 485/sqrt(14) = 129.6217 rounded to 4 decimal places.
degrees of freedom = 14 minus 1 = 13.
t-score = (2815 - 3025)/129.6217 = -1.62010 rounded to 4 decimal places.
you need to know the confidence level.
at 95% confidence level, the two tailed critical t-score with 13 degrees of freedom = plus or minus 2.1604 rounded to 4 decimal places.
since the absolute value of the test t-score is less than the absolute value of the critical t-score, there is not enough evidence to say that the population mean has changed.
the other way to look at it is to look at the test alpha versus the critical alpha.
the critical alpha is .025.
the test alpha is the area to the left of the t-score of -1.62010.
that is equal to .06460 rounded to 4 decimal places.
that is greater than the critical alpha, therefore the results are not significant, meaning you don't have enough data to indicate that the population mean has changed.
the analysis suggests that the sample mean is consistent with the variation in sample means of the same size, therefore to be expected.
if it doesn't pass at 95% confidence level, then it won't pass at any other test criteria greater than 95%.
in fact, it won't pass at any critical alpha less than .06460.
most test are more stringent than that.
90% confidence interval has critical alpha at .05.
this is still less than the test alpha.
there are some wrinkles having to do with standard deviation of sample rather than of population.
since this was already given, the assumption is that it was calculated correctly.
worst case scenario is that it wasn't.
it doesn't appear that doing it correctly, assuming that it was not done correctly, would change the results any, so there is no reason to go back and make a correction.
here's a reference on standard deviation for population versus standard deviation for sample of given size.
sample versus population standard deviation calculations
with a standard devition of 485, the variance is 485^2 = 235225.
the sum of squares, presuming they used 14 as the divisor, is 235225*14 = 3293150.
divide that by 13, as recommended for sample size of 14, you get a variance of 253319.2309.
take the square root of that to get a standard deviation of 503.30286.
divide that by square root of sample size to get standard error (s) of 134.5.
test t-score at 13 degrees of freedom becomes (2815 - 3025) / 134.5 = -1.56 rounded to 2 decimal places.
still not significant to claim that the population mean has changed.
a reference for a one sample t-test is shown below:
https://www.jmp.com/en_us/statistics-knowledge-portal/t-test/one-sample-t-test.html
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