Sovi.AI - AI Math Tutor

Scan to solve math questions
ENFRDEESJAZH-CNZH-TW

QUESTION IMAGE

question 1 of 6, step 3 of 3 correct an sat prep course claims to incre…

Question

question 1 of 6, step 3 of 3
correct
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

before prep course142013801320120011801050118012801240
after prep course160015201550122013101220123014501250

copy data

step 3 of 3: draw a conclusion and interpret the decision.

answer

tables keypad keyboard shortcuts

  • we reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to support the claim that the sat prep course increases student scores by more than 60 points on average.
  • we reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance to support the claim that the sat prep course increases student scores by more than 60 points on average.
  • we fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to support the claim that the sat prep course increases student scores by more than 60 points on average.

Explanation:

Brief Explanations

To determine the conclusion, we first need to recall the hypothesis test for paired differences. The claim is that the prep course increases scores by more than 60 points on average, so the alternative hypothesis is \( H_1: \mu_d> 60 \) and the null hypothesis is \( H_0:\mu_d\leq60 \).

First, we calculate the paired differences \( d_i=\text{After}-\text{Before} \) for each student:

  • For the first student: \( 1600 - 1420=180 \)
  • Second: \( 1520 - 1380 = 140 \)
  • Third: \( 1550 - 1320=230 \)
  • Fourth: \( 1220 - 1200 = 20 \)
  • Fifth: \( 1310 - 1180=130 \)
  • Sixth: \( 1220 - 1050 = 170 \)
  • Seventh: \( 1230 - 1180=50 \)
  • Eighth: \( 1450 - 1280 = 170 \)
  • Ninth: \( 1250 - 1240=10 \)

Next, we calculate the sample mean \( \bar{d} \) and sample standard deviation \( s_d \) of these differences.

The sum of the differences \( \sum d_i=180 + 140+230 + 20+130+170+50+170+10=1000 \)

The sample size \( n = 9 \), so the sample mean \( \bar{d}=\frac{\sum d_i}{n}=\frac{1000}{9}\approx111.11 \)

To calculate the sample standard deviation, we first find the sum of squared deviations from the mean.

\( (180 - 111.11)^2+(140 - 111.11)^2+(230 - 111.11)^2+(20 - 111.11)^2+(130 - 111.11)^2+(170 - 111.11)^2+(50 - 111.11)^2+(170 - 111.11)^2+(10 - 111.11)^2 \)

Calculating each term:

  • \( (68.89)^2\approx4745.83 \)
  • \( (28.89)^2\approx834.63 \)
  • \( (118.89)^2\approx14135.83 \)
  • \( (- 91.11)^2\approx8301.03 \)
  • \( (18.89)^2\approx356.83 \)
  • \( (58.89)^2\approx3468.03 \)
  • \( (-61.11)^2\approx3734.43 \)
  • \( (58.89)^2\approx3468.03 \)
  • \( (-101.11)^2\approx10223.23 \)

Summing these squared deviations: \( 4745.83+834.63 + 14135.83+8301.03+356.83+3468.03+3734.43+3468.03+10223.23=49267.8 \)

The sample variance \( s_d^2=\frac{\sum (d_i-\bar{d})^2}{n - 1}=\frac{49267.8}{8}\approx6158.48 \)

The sample standard deviation \( s_d=\sqrt{6158.48}\approx78.48 \)

Now, we calculate the t - statistic: \( t=\frac{\bar{d}-\mu_{d_0}}{s_d/\sqrt{n}} \), where \( \mu_{d_0} = 60 \) (under the null hypothesis)

\( t=\frac{111.11 - 60}{78.48/\sqrt{9}}=\frac{51.11}{78.48/3}=\frac{51.11}{26.16}\approx1.95 \)

The degrees of freedom \( df=n - 1=8 \). For a one - tailed test at \( \alpha = 0.01 \), the critical value \( t_{\alpha,df}=t_{0.01,8}=2.896 \) (from t - distribution table)

Since the calculated t - statistic \( t = 1.95<2.896 \) (the critical value), we fail to reject the null hypothesis? Wait, no, wait. Wait, maybe I made a mistake in calculation. Wait, let's recalculate the differences:

Wait, first student: 1600 - 1420 = 180

Second: 1520 - 1380 = 140

Third: 1550 - 1320 = 230

Fourth: 1220 - 1200 = 20

Fifth: 1310 - 1180 = 130

Sixth: 1220 - 1050 = 170

Seventh: 1230 - 1180 = 50

Eighth: 1450 - 1280 = 170

Ninth: 1250 - 1240 = 10

Sum: 180+140 = 320; 320+230 = 550; 550+20 = 570; 570+130 = 700; 700+170 = 870; 870+50 = 920; 920+170 = 1090; 1090+10 = 1100. Oh! I made a mistake in the sum earlier. The correct sum of \( d_i \) is 1100. So \( \bar{d}=\frac{1100}{9}\approx122.22 \)

Now recalculate the squared deviations:

\( (180 - 122.22)^2=(57.78)^2\approx3338.53 \)

\( (140 - 122.22)^2=(17.78)^2\approx316.13 \)

\( (230 - 122.22)^2=(107.78)^2\approx11616.53 \)

\( (20 - 122.22)^2=(- 102.22)^2\approx10448.93 \)

\( (130 - 122.22)^2=(7.78)^2\approx60.53 \)

\( (170 - 122.22)^2=(47.78)^2\approx2282.93 \)

\( (50 - 122.22)^2=(-72.22)^2\approx5215.73 \)

\( (170 - 122.22)^2=(47.78)^2\approx2282.93 \)

\( (10 - 122.22)^2=(-112.22)^2\approx12593.33 \)

Now sum these squared deviations:

3338.53+316.13 = 3654.66; +11616.53 = 15271.19; +10448.93 = 25720.12; +60.53 = 25780.…

Answer:

We reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to support the claim that the SAT prep course increases student scores by more than 60 points on average.