Correction on 9.73 Statistics on Large Sample Tests of Hypotheses?

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SUMMARY

The discussion centers on the statistical analysis of SAT scores for California high school students in 2003 compared to national averages. The sample data indicates a verbal score average of 499 and a math score average of 516, with standard deviations of 98 and 96, respectively. The correct statistical test for part a is a t-test due to the unknown population standard deviation, despite the sample size being large enough to approximate a z-test. The conclusion is that the book's answer of z = -3.33 is incorrect, as the proper calculation yields z = -0.816, which does not provide sufficient evidence to reject the null hypothesis at alpha = 0.05.

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1. Homework Statement
(I forgot to write out the problem statement in previous post)
How do California H.S. students compare to students nationwide in their college readiness, as measured by their SAT scores? The national average scores for the class of 2003 were 507 on the verbal portion and 519 on the math portion. Suppose that 100 California students from the class of 2003 were randomly selected and their SAT scores recorded as:

Verbal: Sample average = 499, Sample standard dev = 98

Math: Sample average = 516, Sample standard dev = 96

a. Do the data provide sufficient evidence to indicate that the average verbal score for all California students in the class of 2003 different from the national average? Test using alpha = .05.is different from the national average? Test using alpha = .05.

b. Do the data provide sufficient evidence to indicate that the average math score for all California students in the class of 2003 is different from the national average? Test using alpha = .05.

c. Could you use this data to determine if there is a difference between the average math and verbal scores for all California
students in the class of 2003? Explain your answer.

2. Homework Equations
Relevant DATA:
n = 100 (California students sampled)
national average scores for verbal = 507
sample California average for verbal = 499
standard dev California for verbal = 98

Relevant EQUATION:

SE = standard dev/sqrt(n)
z = [sample California average - mean National]
[standard dev Californial for verbal/sqrt(n)]
OR
z = [sample California average - mean National]/ SE

3. The Attempt at a Solution
Since, standard dev of california sample for verbal test = 98 and n = 100 samples
SE = standard dev/sqrt(n) = 98/sqrt(100) = 9.8
Thus z = [mean of sample California - mean of National (Popn)]/SE
= [499-507]/9.8
= -8/9.8
= - 0.816

This answer for part a. is wrong.

Answers at the back of the book:
a. yes; z = -3.33 b. no; z = 1 c. no
I am stuck at part a. My wrong answer is z = -0.816. The back of the book says z = -3.33. Please show me how step-by-step to derive at the correct answer. Thank you very much.

I get the same answer of z = -0.816 instead of -3.33. I am only concerned about getting the z value because how can I answer yes or no, without knowing it.
Also, it is asking for us to test using alpha = .05. Z-test is the only test that seems to apply here since we were are comparing a sample (California) from a population (National). The textbook is expecting students to know which tests (z-test, t-test, etc) to solve the problem. Also, the back of the book even says z = -3.33
 
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Book's answer for part a is wrong on two accounts. You should use a t test; and the answer isn't yes. The reason for using a t test is because the standard deviation is not known, but estimated from the sample. In this case, the "z" statistic in fact follows a t distribution. However, with a large sample size (e.g. n > 30), the t distribution becomes close enough to the standard normal distribution that the z test can be used.
 

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