In the population of students who take the SAT, the average combined score is 1600. I wonder whether students in Ohio differ from the population of students who take the SAT. Suppose I select a random sample of 100 students from Ohio and find that their average SAT score was 1630 with a standard deviation of 200.
a. Write the null and alternative hypotheses for this research question.
b. Is the difference between the sample mean and the population mean is statistically significant using an alpha level of .05?
c. Then calculate the 99% confidence interval for the mean and wrap words around it.

Question

Quizplus · Accepted Answer

The Answer is a. Null hypothesis: The sample mean for Ohio students equals the population mean on the SAT.
Alternative hypothesis: The sample mean for Ohio students does not equal the population mean.
b.First, we must first calculate a standard error of the mean:

S_{\bar{x}}

=

\frac{200}{\sqrt{100}}

= 200/10 = 20
Next, calculate the t value for this problem and determine whether it is statistically significant.
First of all, we do not know the population standard deviation, so we have to calculate a t value, not a z score. To do this, we must first calculate a standard error of the mean:
Here is how I can calculate a t value: t =

\frac{1630-1600}{20}

= 1.5
The degrees of freedom for this problem is n - 1, so 100 - 1 = 99. Now look in Appendix B. Notice that 99 is between 120 and 60. When we have a situation like this, go with the smaller df in Appendix B, unless your df is much closer to one of the values in Appendix B than the other. So let's take a look at the row with 60 degrees of freedom in Appendix B. We are looking for the value closest to our t value of 1.5. We can see that our observed t value of 1.5 is between 1.296 and 1.671 for df = 60. Go up to the top of these two columns to where it says ..20 and .10. That is the probability of getting a t value of 1.5 by chance with 60 degrees of freedom. Out t value is 1.5, so the probability of getting this value by chance is between .10 and .20. This is LARGER THAN our alpha level of .05, so we will conclude the difference between our sample and population means was NOT statistically significant. Ohio student did not do better, on average, than the population of SAT takers.
c. CI = sample mean +/- (t)(

S_{\bar{x}}

)
We already know the sample mean (1630) and the standard error of the mean (20). So all we need to do is look up the t value in Appendix B. We are still using 60 degrees of freedom and we are looking for the column that says .01 for 2-tailed test. We are using .01 here because our confidence interval is a 99% CI, and 1-.99 = .01. So using this information we find a t value of 2.660. Now we do the math.
CI₉₉ = 1630 +/- (2.660) (20)

\rightarrow

1630 +/- 53.20. So our confidence interval is 1576.80 to 1683.20. We are 99% confident that the population mean is between 1576.80 and 1683.20.

In the Population of Students Who Take the SAT, the Average