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 0.10 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	1680	1130	1230	1240	1260	1600	1310	1700	1290

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

answer

Explanation:

Step1: Calculate the differences \( d \)

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

• \( 1680 - 1500 = 180 \)
• \( 1130 - 1090 = 40 \)
• \( 1230 - 1140 = 90 \)
• \( 1240 - 1200 = 40 \)
• \( 1260 - 1040 = 220 \)
• \( 1600 - 1350 = 250 \)
• \( 1310 - 1210 = 100 \)
• \( 1700 - 1500 = 200 \)
• \( 1290 - 1210 = 80 \)

So the differences \( d \) are: \( 180, 40, 90, 40, 220, 250, 100, 200, 80 \)

Step2: Calculate the mean of \( d \) (\( \bar{d} \))

The formula for the mean is \( \bar{d} = \frac{\sum d}{n} \), where \( n = 9 \).
First, sum the differences:
\( 180 + 40 + 90 + 40 + 220 + 250 + 100 + 200 + 80 = 1200 \)
Then, \( \bar{d} = \frac{1200}{9} \approx 133.3333 \)

Step3: Calculate the standard deviation of \( d \) (\( s_d \))

First, calculate the squared differences from the mean:

• \( (180 - 133.3333)^2 \approx (46.6667)^2 \approx 2177.7778 \)
• \( (40 - 133.3333)^2 \approx (-93.3333)^2 \approx 8711.1111 \)
• \( (90 - 133.3333)^2 \approx (-43.3333)^2 \approx 1877.7778 \)
• \( (40 - 133.3333)^2 \approx (-93.3333)^2 \approx 8711.1111 \)
• \( (220 - 133.3333)^2 \approx (86.6667)^2 \approx 7511.1111 \)
• \( (250 - 133.3333)^2 \approx (116.6667)^2 \approx 13611.1111 \)
• \( (100 - 133.3333)^2 \approx (-33.3333)^2 \approx 1111.1111 \)
• \( (200 - 133.3333)^2 \approx (66.6667)^2 \approx 4444.4444 \)
• \( (80 - 133.3333)^2 \approx (-53.3333)^2 \approx 2844.4444 \)

Sum of squared differences:
\( 2177.7778 + 8711.1111 + 1877.7778 + 8711.1111 + 7511.1111 + 13611.1111 + 1111.1111 + 4444.4444 + 2844.4444 = 41000 \)

The sample variance \( s_d^2 = \frac{\sum (d - \bar{d})^2}{n - 1} = \frac{41000}{8} = 5125 \)
The sample standard deviation \( s_d = \sqrt{5125} \approx 71.5892 \)

Step4: Calculate the test statistic (t - statistic)

The formula for the t - statistic in a paired t - test is \( t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}} \)
We are testing the claim that the average increase is more than 60 points, so \( \mu_d = 60 \)
Substitute the values:
\( t=\frac{133.3333 - 60}{71.5892/\sqrt{9}}=\frac{73.3333}{71.5892/3}=\frac{73.3333}{23.8631}\approx3.073 \)

Answer:

\( 3.073 \)