About 130000 high school students took the AP exam in 2010 .the free response section of the exam consisted of five open ended p
1 answer:
Complete question is;
About 130,000 high school students took the AP Statistics exam in 2010. The free-response section of the exam consisted of five open-ended problems and an investigative task. Each free-response question is scored on a 0 to 4 scale (with 4 being the best). For one of the problems, a random sample of 30 student papers yielded the scores that are graphed in the dot plot attached. The mean score for this sample is x¯ = 1.267 and the standard deviation is s = 1.230.
(a) Find and interpret the standard error of the mean.
(b) Construct and interpret a 99% confidence interval to estimate the mean score on this question. Use the four-step process.
Answer:
A) SOE = 0.225
The interpretation of this is that: if a sample size of 30 was taken, the difference between a sample mean score and a population mean score will be an average of 0.225
B) CI = (0.648, 1.886)
The interpretation of this is that;
We are 99% confident that that interval (0.648, 1.886) captures the true mean score of the AP Statistics exam in 2010
Step-by-step explanation:
A) We are given;
sample mean; x¯ = 1.267
Sample standard deviation; s = 1.230
Sample size; n = 30
Formula for standard error of the mean is;
SOE = s/√n
SOE = 1.23/√30
SOE = 0.225
The interpretation of this is that: if a sample size of 30 was taken, the difference between a sample mean score and a population mean score will be an average of 0.225.
B) we have a sample size of n = 30.
Thus;
DF = 30 - 1 = 29
We want to do a 99% Confidence interval.
From the attached t-table, we can see that critical t-value for DF = 29 and CL of 99%, we have t = 2.756
Formula for CI is;
CI = x¯ ± t(s/√n)
CI = 1.267 ± 2.756(1.23/√30)
CI = 1.267 ± 0.619
CI = (0.648,1.886)
The interpretation of this is that;
We are 99% confident that that interval (0.648, 1.886) captures the true mean score of the AP Statistics exam in 2010.
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