Please explain how you got the answer. Don’t just tell me the answer
1 answer:
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Answer:
The test statistic needed to evaluate the claim is t = -1.08.
Step-by-step explanation:
The test statistic is:
In which X is the sample mean, is the expected value of the mean, s is the standard deviation of the sample and n is the size of the sample.
At a certain university, the average attendance at basketball games has been 3125. The athletic director claims that the attendance is the same as last year.
This means that
Due to the dismal showing of the team this year, the attendance for the first 9 games has averaged only 2915 with a standard deviation of 585.
This means that
What is the test statistic needed to evaluate the claim?
The test statistic needed to evaluate the claim is t = -1.08.
I got D.
There's a few ways to solve it; I prefer using tables, but there are functions on a TI-84 that'll do it for you too. The logic here is, you have a standard normal distribution which means right away, the mean is 0 and the standard deviation is 1. This means you can use a Z table that helps you calculate the area beneath a normal curve for a range of values. Here, your two Z scores are -1.21 and .84. You might notice that this table doesn't account for negative values, but the cool thing about a normal distribution is that we can assume symmetry, so you can just look for 1.21 and call it good. The actual calculation here is:
1 - Z-score of 1.21 - Z-score .84 ... use the table or calculator
1 - .1131 - .2005 = .6864
Because this table calculates areas to the RIGHT of the mean, you have to play around with it a little to get the bit in the middle that your graph asks for. You subtract from 1 to make sure you're getting the area in the middle and not the area of the tails in this problem.
