Coaching companies claim that their courses can raise the SAT scores of high school students. But students who retake the SAT without paying for coaching also usually raise their scores. A random sample of students who took the SAT twice found 427 who were coached and 2733 who were uncoached. Starting with their verbal scores on the first and second tries, we have these summary statistics:
Try 1 Try2 Gain
n xbar s xbar s xbar s
Coached 427 500 92 529 97 29 59
Uncoached 2733 506 101 527 101 21 52
1)Use Table C to estimate a 99% confidence interval for the mean gain of all students who are coached.
_________to________at 99% confidence.
2) (a) Give the alternative hypothesis: μ1−μ2 > 0.
(b) Give the t test statistic: 2.6459
(c) Give the appropriate critical value for α=5%:_____
I got #2 part a and b right, but I need help finding (#1) the confidence interval and (c) the critical value for α=5%.
To find the confidence interval for the mean gain of all students who are coached, you can use the following formula:
Confidence Interval = (xbar1 - xbar2) ± t * sqrt((s1^2 / n1) + (s2^2 / n2))
Where:
xbar1 = sample mean of the gain for the coached students
xbar2 = sample mean of the gain for the uncoached students
s1 = sample standard deviation of the gain for the coached students
s2 = sample standard deviation of the gain for the uncoached students
n1 = number of coached students
n2 = number of uncoached students
t = critical value from the t-distribution with df = n1 + n2 - 2 at the desired confidence level
Let's calculate the confidence interval using the given information:
From the table, we have:
xbar1 = 29
xbar2 = 21
s1 = 59
s2 = 52
n1 = 427
n2 = 2733
First, let's calculate the critical value using Table C for a 99% confidence level:
df = n1 + n2 - 2 = 427 + 2733 - 2 = 3158
t-critical value for α = 0.01 (99% confidence level) and df = 3158 is approximately 2.607.
Now, we can plug in the values into the confidence interval formula:
Confidence Interval = (29 - 21) ± 2.607 * sqrt((59^2 / 427) + (52^2 / 2733))
Confidence Interval = 8 ± 2.607 * sqrt((3481/ 427) + (2704 / 2733))
Confidence Interval = 8 ± 2.607 * sqrt(8.147 + 0.990)
Confidence Interval = 8 ± 2.607 * sqrt(9.137)
Confidence Interval = 8 ± 2.607 * 3.025
Confidence Interval = 8 ± 7.886
Therefore, the 99% confidence interval for the mean gain of all students who are coached is from 8 - 7.886 to 8 + 7.886, which is approximately (-0.886, 16.886).
Regarding (c) the appropriate critical value for α=5%, you can find it by subtracting 0.05 (5%) from 1 and finding the corresponding value in the t-distribution table with the degrees of freedom (df) equal to the total sample size minus 2 (3158). Using table C, the appropriate critical value for α=5% is approximately 1.960.
To find the 99% confidence interval for the mean gain of all students who are coached, we need to use Table C, which provides critical values for t-distributions. The formula to calculate the confidence interval is:
Confidence Interval = x̄ ± (t* × (s/√n))
Where:
x̄ = sample mean of the gain for coached students
s = sample standard deviation of the gain for coached students
n = sample size of coached students
t* = critical value based on the desired confidence level and degrees of freedom
Looking at the information given, we see that for the coached students:
x̄ = 29
s = 59
n = 427
To find the critical value, we need the degrees of freedom, which is calculated as (sample size - 1) for each sample. Since we are only given the sample size of the coached students, we use their sample size minus 1.
Degrees of freedom = n - 1 = 426
To find the critical value for a 99% confidence level, we need to find the value in Table C that corresponds to a 99% confidence level and 426 degrees of freedom. This can be done by comparing the confidence level with the probability in the table. In this case, we want to find the value closest to a 99% confidence level, which is greater than 99%.
Once we find the critical value, we plug in all the values into the formula to calculate the confidence interval.
For part (c) - to determine the appropriate critical value for α = 5%, we need to find a value in Table C that corresponds to a significance level (α) of 0.05. The critical value for a two-tailed test would be determined by splitting the alpha value equally into the two tails (0.025 in each tail). We would then look for the value in the table that is closest to 0.025, given the degrees of freedom for the t-distribution.
Unfortunately, I don't have access to Table C as an AI. However, you can find Table C in most statistics textbooks or search for it online to determine the correct critical value for a 99% confidence level and α=5%.
z = 2.575 for a 99% confidence. Since your sample sizes are large, I would use a two-sample confidence interval formula for large sample sizes.
Critical value at .05 using a z-table for a one-tailed test is z = 1.645
If you use a t-table instead to find critical value, then you will have to figure degrees of freedom as well.