MAT Scores Analysis for Martian High School Students, Exams of Probability and Statistics

An analysis of mat (martian aptitude test) scores for a sample of martian high school students. The calculation of the mean and standard deviation of the scores, as well as the creation of a stemplot to check for normality. The document also determines the sample size needed to achieve a margin of error of 0.5.

Typology: Exams

Pre 2010

Uploaded on 07/22/2009

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Worksheet 5 9/03/08
Problem 1. Before they attend college, every Martian takes the MAT (Martian Aptitude
Test). An SRS of Martian high school students is collected, and the results are below:
33.2 31.9 32.6 26.5 33.3
32.3 33.0 32.0 30.5 32.7
23.0 30.9 32.7 33.7 32.3
24.1 30.2 31.3 28.7 31.9
The population standard deviation of MAT scores is σ= 3.
(a) We know for the SAT, that the scores are approximately normal. From this sample,
does it appear that MAT scores are also approximately Normally distributed? Make a
stemplot to check.
23 |0
24 |1
25 |
26 |5
27 |
28 |7
29 |
30 |259
31 |399
32 |033677
33 |0237
Actually, appears a bit left skewed. (However, the overall distribution may be Normal)
(b) Assuming that it is in fact Normal, calculate a 90% confidence interval for the mean
MAT score.
Calculating, we get ¯x= 30.84. For 90% confidence z= 1.645. There are 20 scores, so
our confidence interval is
30.84 ±1.645 3
20 = 30.84 ±1.10 = (29.74,31.94)
(c) What size sample do we need to collect to get the margin of error down to .5?
We use the formula n=zσ
m2where mis margin of error.
n=(1.645)(3)
.52
= 97.417
Since we can’t have a partial person, we round this up to 98.

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Worksheet 5 9/03/

Problem 1. Before they attend college, every Martian takes the MAT (Martian Aptitude Test). An SRS of Martian high school students is collected, and the results are below:

  1. 2 31. 9 32. 6 26. 5 33. 3
  2. 3 33. 0 32. 0 30. 5 32. 7
  3. 0 30. 9 32. 7 33. 7 32. 3
  4. 1 30. 2 31. 3 28. 7 31. 9

The population standard deviation of MAT scores is σ = 3. (a) We know for the SAT, that the scores are approximately normal. From this sample, does it appear that MAT scores are also approximately Normally distributed? Make a stemplot to check.

23 | 0 24 | 1 25 | 26 | 5 27 | 28 | 7 29 | 30 | 259 31 | 399 32 | 033677 33 | 0237 Actually, appears a bit left skewed. (However, the overall distribution may be Normal) (b) Assuming that it is in fact Normal, calculate a 90% confidence interval for the mean MAT score.

Calculating, we get ¯x = 30.84. For 90% confidence z∗^ = 1.645. There are 20 scores, so our confidence interval is

  1. 84 ± 1. 645

(c) What size sample do we need to collect to get the margin of error down to .5?

We use the formula n =

(z∗σ m

where m is margin of error.

n =

Since we can’t have a partial person, we round this up to 98.