Solved on Nov 24, 2023

Test claim that SAT prep course increases scores by >60 points on avg. 9 students, $H_0$ : $\mu_1 - \mu_2 \leq 60$ , $\alpha = 0.10$ , normal distribution.

STEP 1

Assumptions1. The scores before the prep course are from Population1. The scores after the prep course are from Population3. The claim is that the prep course increases scores by more than60 points on average4. The level of significance is0.105. The population distribution of the paired differences is approximately normal6. There are9 students who have taken the prep course

STEP 2

First, we need to calculate the differences in scores for each student. This can be done by subtracting the score before the prep course (Population1) from the score after the prep course (Population2).
$Difference = Score_{after} - Score_{before}$

STEP 3

Calculate the differences for each student using the given scores.

STEP 4

Now, we need to calculate the mean of these differences. This can be done by adding up all the differences and dividing by the number of students.
$Mean_{difference} = \frac{\sum Difference}{n}$

STEP 5

Calculate the mean difference using the calculated differences and the number of students (9).

STEP 6

Next, we need to calculate the standard deviation of the differences. This can be done using the formula $SD_{difference} = \sqrt{\frac{\sum (Difference - Mean_{difference})^2}{n-1}}$

STEP 7

Calculate the standard deviation of the differences using the calculated differences, the mean difference, and the number of students (9).

STEP 8

Now, we can calculate the test statistic. This is done by subtracting the claimed mean difference (60) from the calculated mean difference and dividing by the standard error of the mean difference. The standard error can be calculated as the standard deviation divided by the square root of the number of students.
$Test\, statistic = \frac{Mean_{difference} - Claimed\, mean}{SD_{difference}/\sqrt{n}}$

STEP 9

Calculate the test statistic using the calculated mean difference, the claimed mean difference (60), the calculated standard deviation of the differences, and the number of students (9).

STEP 10

Now, we need to find the critical value for a one-tailed test at a0.10 level of significance. This can be found from a t-distribution table with degrees of freedom equal to the number of students minus.

STEP 11

Compare the calculated test statistic with the critical value. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is sufficient evidence at a0.10 level of significance to support the claim that the SAT prep course increases student scores by more than60 points on average. If the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis and conclude that there is insufficient evidence at a0.10 level of significance to support the claim.

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Test claim that SAT prep course increases scores by >60 points on avg. 9 students, $H_0$ : $\mu_1 - \mu_2 \leq 60$ , $\alpha = 0.10$ , normal distribution.

STEP 1

STEP 2

First, we need to calculate the differences in scores for each student. This can be done by subtracting the score before the prep course (Population1) from the score after the prep course (Population2).
$Difference = Score_{after} - Score_{before}$

STEP 3

Calculate the differences for each student using the given scores.

STEP 4

Now, we need to calculate the mean of these differences. This can be done by adding up all the differences and dividing by the number of students.
$Mean_{difference} = \frac{\sum Difference}{n}$

STEP 5

Calculate the mean difference using the calculated differences and the number of students (9).

STEP 6

Next, we need to calculate the standard deviation of the differences. This can be done using the formula $SD_{difference} = \sqrt{\frac{\sum (Difference - Mean_{difference})^2}{n-1}}$

STEP 7

Calculate the standard deviation of the differences using the calculated differences, the mean difference, and the number of students (9).

STEP 8

STEP 9

Calculate the test statistic using the calculated mean difference, the claimed mean difference (60), the calculated standard deviation of the differences, and the number of students (9).

STEP 10

Now, we need to find the critical value for a one-tailed test at a0.10 level of significance. This can be found from a t-distribution table with degrees of freedom equal to the number of students minus.

STEP 11

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Test claim that SAT prep course increases scores by >60 points on avg. 9 students, H0H_0H0​: μ1−μ2≤60\mu_1 - \mu_2 \leq 60μ1​−μ2​≤60, α=0.10\alpha = 0.10α=0.10, normal distribution.

STEP 1

STEP 2

STEP 3

Calculate the differences for each student using the given scores.

STEP 4

Now, we need to calculate the mean of these differences. This can be done by adding up all the differences and dividing by the number of students.Meandifference=∑DifferencenMean_{difference} = \frac{\sum Difference}{n}Meandifference​=n∑Difference​

STEP 5

Calculate the mean difference using the calculated differences and the number of students (9).

STEP 6

Next, we need to calculate the standard deviation of the differences. This can be done using the formulaSDdifference=∑(Difference−Meandifference)2n−1SD_{difference} = \sqrt{\frac{\sum (Difference - Mean_{difference})^2}{n-1}}SDdifference​=n−1∑(Difference−Meandifference​)2​​

STEP 7

Calculate the standard deviation of the differences using the calculated differences, the mean difference, and the number of students (9).

STEP 8

STEP 9

Calculate the test statistic using the calculated mean difference, the claimed mean difference (60), the calculated standard deviation of the differences, and the number of students (9).

STEP 10

Now, we need to find the critical value for a one-tailed test at a0.10 level of significance. This can be found from a t-distribution table with degrees of freedom equal to the number of students minus.

STEP 11

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Test claim that SAT prep course increases scores by >60 points on avg. 9 students, $H_0$ : $\mu_1 - \mu_2 \leq 60$ , $\alpha = 0.10$ , normal distribution.

Now, we need to calculate the mean of these differences. This can be done by adding up all the differences and dividing by the number of students.
$Mean_{difference} = \frac{\sum Difference}{n}$

Next, we need to calculate the standard deviation of the differences. This can be done using the formula $SD_{difference} = \sqrt{\frac{\sum (Difference - Mean_{difference})^2}{n-1}}$