9.73 Statistics on Large Sample Tests of Hypotheses?

[Little Bear]

Homework Statement

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 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.

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

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.

 

[mighty2000]

So why are we using t-test?

it is important to carefully analyze and interpret data in order to draw accurate conclusions. In this case, we are comparing the average SAT scores of California high school students to the national average for the class of 2003. To determine if there is a significant difference between the two, we need to use a statistical test.

In this problem, we are given the sample averages and standard deviations for both the verbal and math portions of the SAT. We also know the national average scores for the class of 2003. We can use a z-test to determine if there is a significant difference between the sample means and the population mean.

a. To test if the average verbal score for all California students is different from the national average, we can use the following formula:

z = (sample mean - population mean) / (standard deviation / √n)

Plugging in the given values, we get:

z = (499 - 507) / (98 / √100) = -0.816

To determine if this result is statistically significant, we need to compare it to the critical value at a 0.05 significance level. Looking at a z-table, we can see that the critical value for a one-tailed test at α = 0.05 is -1.645. Since our calculated z-value is greater than the critical value, we can reject the null hypothesis and conclude that there is a significant difference between the average verbal score of California students and the national average.

b. To test if the average math score for all California students is different from the national average, we can use the same formula as above:

z = (sample mean - population mean) / (standard deviation / √n)

Plugging in the given values, we get:

z = (516 - 519) / (96 / √100) = -0.3125

Again, we compare this result to the critical value at a 0.05 significance level. The critical value for a one-tailed test at α = 0.05 is -1.645, which is greater than our calculated z-value. Therefore, we fail to reject the null hypothesis and conclude that there is not a significant difference between the average math score of California students and the national average.

c. We cannot use this data to determine if there is a difference between the average math and verbal scores for all California students