Question

an sat prep course claims to increase student scores by more than 60 points, on average. to test this claim, 9 students who have previously taken the sat are randomly chosen to take the prep course. their sat scores before and after completing the prep course are listed in the following table. test the claim at the 0.01 level of significance assuming that the population distribution of the paired differences is approximately normal. let ( d = ) (scores after completing the prep course) ( - ) (scores before completing the prep course).

sat scores
after prep course	1600	1520	1550	1220	1310	1220	1230	1450	1250

step 2 of 3 : compute the value of the test statistic. round your answer to three decimal places.

answer

Explanation:

Step 1: Calculate the differences \( d \)

First, we calculate the difference \( d \) for each student as \( d = \text{After} - \text{Before} \).
For each pair:

• \( 1600 - 1420 = 180 \)
• \( 1520 - 1380 = 140 \)
• \( 1550 - 1320 = 230 \)
• \( 1220 - 1200 = 20 \)
• \( 1310 - 1180 = 130 \)
• \( 1220 - 1050 = 170 \)
• \( 1230 - 1180 = 50 \)
• \( 1450 - 1280 = 170 \)
• \( 1250 - 1240 = 10 \)

So the differences \( d \) are: \( 180, 140, 230, 20, 130, 170, 50, 170, 10 \)

Step 2: Calculate the mean of \( d \) (\( \bar{d} \))

The formula for the mean \( \bar{d} \) is \( \bar{d} = \frac{\sum d}{n} \), where \( n = 9 \).
First, sum the differences:
\( 180 + 140 + 230 + 20 + 130 + 170 + 50 + 170 + 10 = 1000 \)
Then, \( \bar{d} = \frac{1000}{9} \approx 111.111 \)

Step 3: Calculate the standard deviation of \( d \) (\( s_d \))

First, find the squared differences from the mean:

• \( (180 - 111.111)^2 \approx 4722.222 \)
• \( (140 - 111.111)^2 \approx 834.012 \)
• \( (230 - 111.111)^2 \approx 14135.802 \)
• \( (20 - 111.111)^2 \approx 8200.012 \)
• \( (130 - 111.111)^2 \approx 356.012 \)
• \( (170 - 111.111)^2 \approx 3467.902 \)
• \( (50 - 111.111)^2 \approx 3734.012 \)
• \( (170 - 111.111)^2 \approx 3467.902 \)
• \( (10 - 111.111)^2 \approx 10224.012 \)

Sum these squared differences:
\( 4722.222 + 834.012 + 14135.802 + 8200.012 + 356.012 + 3467.902 + 3734.012 + 3467.902 + 10224.012 = 49141.878 \)

The sample variance \( s_d^2 = \frac{\sum (d - \bar{d})^2}{n - 1} = \frac{49141.878}{8} \approx 6142.735 \)

The sample standard deviation \( s_d = \sqrt{6142.735} \approx 78.376 \)

Step 4: Calculate the test statistic \( t \)

The formula for the test statistic in a paired t - test is \( t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}} \), where \( \mu_d = 60 \) (the claimed mean difference), \( \bar{d}\approx111.111 \), \( s_d\approx78.376 \), and \( n = 9 \).

First, calculate the denominator: \( s_d/\sqrt{n}=\frac{78.376}{\sqrt{9}}=\frac{78.376}{3}\approx26.125 \)

Then, calculate the numerator: \( \bar{d}-\mu_d = 111.111 - 60 = 51.111 \)

Now, calculate the test statistic: \( t=\frac{51.111}{26.125}\approx1.956 \)

Answer:

\( 1.956 \)