A sample of 10 students took a class designed to improve their SAT math scores. Following are their scores before and after the class. Can you conclude that the mean increase in scores is greater than 15 points? Let represent the mean score after the class and uvd=uv1-uv2. Use the a=0.10 level and the P-value method.

Question

Before:
394
483
447
440
439
526
473
440
481
459

After:
430
482
482
479
451
529
493
466
482
455

(a) State the appropriate null and alternate hypotheses.

Answer 1

To analyze whether the mean increase in SAT math scores after the class is greater than 15 points, we need to formulate our null and alternate hypotheses based on the mean differences between the before and after scores.

Let \( \mu_d \) represent the mean difference in scores (after - before).

The null and alternative hypotheses are as follows:

• Null Hypothesis (\( H_0 \)): The mean increase in scores is less than or equal to 15 points.

  \( H_0: \mu_d \leq 15 \)

• Alternative Hypothesis (\( H_a \)): The mean increase in scores is greater than 15 points.

  \( H_a: \mu_d > 15 \)

Summary of Hypotheses:

• \( H_0: \mu_d \leq 15 \)
• \( H_a: \mu_d > 15 \)

Answer 2

IS THIS A TWO TAILED test?

Answer 3

No, this is not a two-tailed test; it is a one-tailed test.

In the hypothesis formulation:

• The null hypothesis (\( H_0 \)) states that the mean increase in scores is less than or equal to 15 points.
• The alternative hypothesis (\( H_a \)) asserts that the mean increase in scores is greater than 15 points.

Since the alternative hypothesis is specifically looking for an increase (greater than 15), it indicates one direction (the right tail) of the distribution, thus making it a one-tailed test.

In summary:

• This is a one-tailed test because we are testing specifically for increases in means rather than differences in both directions.

Answer 4

(b) Compute the P-value. Round the answer to at least four decimal places.

Answer 5

To compute the P-value, we first need to find the differences in scores for each student (after - before), compute the mean and standard deviation of those differences, and then perform a one-sample t-test. Here's how to do it:

Step 1: Calculate the Differences

The scores before and after the class are given as follows:

Before:

• 394
• 483
• 447
• 440
• 439
• 526
• 473
• 440
• 481
• 459

After:

• 430
• 482
• 482
• 479
• 451
• 529
• 493
• 466
• 482
• 455

Differences (After - Before):

\[ \begin{align*} 430 - 394 & = 36 \ 482 - 483 & = -1 \ 482 - 447 & = 35 \ 479 - 440 & = 39 \ 451 - 439 & = 12 \ 529 - 526 & = 3 \ 493 - 473 & = 20 \ 466 - 440 & = 26 \ 482 - 481 & = 1 \ 455 - 459 & = -4 \ \end{align*} \]

Step 2: Create the differences list

The differences are:

• 36, -1, 35, 39, 12, 3, 20, 26, 1, -4

Step 3: Calculate the mean and standard deviation of the differences

1. Mean of differences (\( \bar{d} \)): \[ \bar{d} = \frac{(36 + (-1) + 35 + 39 + 12 + 3 + 20 + 26 + 1 + (-4))}{10} = \frac{ 36 - 1 + 35 + 39 + 12 + 3 + 20 + 26 + 1 - 4 }{10} = \frac{ 167 }{10} = 16.7 \]

2. Standard deviation of differences (\( s_d \)): \[ s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}} \] Where \( d_i \) are the differences, \( \bar{d} = 16.7 \), and \( n = 10 \).

   • Differences from the mean: \[ \begin{align*} 36 - 16.7 & \approx 19.3 \ -1 - 16.7 & \approx -17.7 \ 35 - 16.7 & \approx 18.3 \ 39 - 16.7 & \approx 22.3 \ 12 - 16.7 & \approx -4.7 \ 3 - 16.7 & \approx -13.7 \ 20 - 16.7 & \approx 3.3 \ 26 - 16.7 & \approx 9.3 \ 1 - 16.7 & \approx -15.7 \ -4 - 16.7 & \approx -20.7 \ \end{align*} \]

   • Squared differences: \[ \begin{align*} (19.3)^2 & \approx 372.49 \ (-17.7)^2 & \approx 313.29 \ (18.3)^2 & \approx 334.89 \ (22.3)^2 & \approx 497.29 \ (-4.7)^2 & \approx 22.09 \ (-13.7)^2 & \approx 187.69 \ (3.3)^2 & \approx 10.89 \ (9.3)^2 & \approx 86.49 \ (-15.7)^2 & \approx 246.49 \ (-20.7)^2 & \approx 428.49 \ \end{align*} \]

   • Sum of squared differences: \[ 372.49 + 313.29 + 334.89 + 497.29 + 22.09 + 187.69 + 10.89 + 86.49 + 246.49 + 428.49 = 2079.4 \]

   • Then: \[ s_d = \sqrt{\frac{2079.4}{10 - 1}} = \sqrt{\frac{2079.4}{9}} \approx \sqrt{230.99} \approx 15.2 \]

Step 4: Calculate the t-statistic

\[ t = \frac{\bar{d} - \mu_0}{s_d/\sqrt{n}} \]

Where \( \mu_0 = 15 \) (the hypothesized mean increase), \( n = 10 \), \( \bar{d} = 16.7 \), and \( s_d \approx 15.2 \):

\[ t = \frac{16.7 - 15}{15.2/\sqrt{10}} = \frac{1.7}{4.8} \approx 0.3542 \]

Step 5: Find the P-value

We need to look up the t-value in the t-distribution table or use software/calculator for a one-tailed test. The degrees of freedom \( df = n - 1 = 9\).

Using a t-table or calculator for one-tailed tests, we find:

• For \( t = 0.3542 \) with \( df = 9 \), the P-value is approximately \( 0.3632 \).

Conclusion

The calculated P-value is approximately:

\[ \text{P-value} \approx 0.3632 \]

Rounded to four decimal places: 0.3632.

Answer 6

dO WE REJECT Hv0?

Answer 7

To determine whether to reject the null hypothesis \( H_0: \mu_d \leq 15 \), we compare the P-value we calculated to the significance level, which is \( \alpha = 0.10 \).

Decision Rule:

• If the P-value is less than or equal to \( \alpha \), reject the null hypothesis.
• If the P-value is greater than \( \alpha \), fail to reject the null hypothesis.

In our case:

• P-value: \( 0.3632 \)
• Significance level: \( \alpha = 0.10 \)

Since \( 0.3632 \) is greater than \( 0.10 \):

• Decision: Fail to reject \( H_0 \).

Conclusion:

There is not enough evidence at the \( \alpha = 0.10 \) level to conclude that the mean increase in SAT math scores after the class is greater than 15 points.