Math / Data & Statistics

QuestionDo students perform the same when they take an exam alone as when they take an exam in a classroom setting? Eight students were given two tests of equal difficulty. They took one test in a solitary room and they took the other in a room filled with other students. The results are shown below.
Exam Scores \begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline Alone & 77 & 80 & 78 & 84 & 89 & 86 & 86 & 67 \\ \hline Classroom & 81 & 83 & 90 & 89 & 92 & 89 & 82 & 69 \\ \hline \end{tabular}
Assume a Normal distribution. What can be concluded at the the $\alpha=0.05$ level of significance level of significance?
For this study, we should use Select an answer a. The null and alternative hypotheses would be: $H_{0}$ : Select an answer Select an answer Select an answer
⓪ (please enter a decimal) $H_{1}$ : Select an answer Select an answer Select an answer (t) (Please enter a decimal) b. The test statistic ? $=$ $\square$ (please show your answer to 3 decimal places.) c. The $p$ -value $=$ $\square$ (Please show your answer to 4 decimal places.) d. The $p$ -value is $\square$ $\alpha$ e. Based on this, we should Select an answer $\square$ $\square$ the null hypothesis. f. Thus, the final conclusion is that ... The results are statistically significant at $\alpha=0.05$ , so there is sufficient evidence to conclude that the eight students scored the same on average taking the exam alone compared to the classroom setting. The results are statistically insignificant at $\alpha=0.05$ , so there is statistically significant evidence to conclude that the population mean test score taking the exam alone is equal to the population mean test score taking the exam in a classroom setting. The results are statistically insignificant at $\alpha=0.05$ , so there is insufficient evidence to conclude that the population mean test score taking the exam alone is not the same as the population mean test score taking the exam in a classroom setting. The results are statistically significant at $\alpha=0.05$ , so there is sufficient evidence to conclude that the population mean test score taking the exam alone is not the same as the population mean test score taking the exam in a classroom setting.

Studdy Solution

STEP 1

1. The exam scores for both conditions (alone and classroom) are normally distributed.
2. The sample size is small (n = 8), so a paired t-test is appropriate.
3. The significance level is $\alpha = 0.05$ .

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the differences between paired scores.
3. Compute the test statistic.
4. Determine the p-value.
5. Compare the p-value with $\alpha$ and make a decision.
6. State the conclusion.

STEP 3

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean difference in scores between the two conditions is zero. - Alternative hypothesis ( $H_1$ ): The mean difference in scores between the two conditions is not zero.

STEP 4

Calculate the differences between paired scores:
\begin{align} \text{Differences} & = (81 - 77), (83 - 80), (90 - 78), (89 - 84), (92 - 89), (89 - 86), (82 - 86), (69 - 67) \\ & = 4, 3, 12, 5, 3, 3, -4, 2 \end{align}

STEP 5

Compute the test statistic:
1. Calculate the mean of differences: $\bar{d} = \frac{4 + 3 + 12 + 5 + 3 + 3 - 4 + 2}{8} = 3.5$
2. Calculate the standard deviation of differences: $s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}$
3. Compute the test statistic: $t = \frac{\bar{d}}{s_d / \sqrt{n}}$

STEP 6

Calculate the standard deviation of differences:
$s_d = \sqrt{\frac{(4-3.5)^2 + (3-3.5)^2 + (12-3.5)^2 + (5-3.5)^2 + (3-3.5)^2 + (3-3.5)^2 + (-4-3.5)^2 + (2-3.5)^2}{8-1}}$
$s_d = \sqrt{\frac{0.25 + 0.25 + 72.25 + 2.25 + 0.25 + 0.25 + 56.25 + 2.25}{7}}$
$s_d = \sqrt{\frac{134}{7}} \approx 4.368$

STEP 7

Compute the test statistic:
$t = \frac{3.5}{4.368 / \sqrt{8}} \approx \frac{3.5}{1.544} \approx 2.267$

STEP 8

Determine the p-value using the t-distribution with $n-1 = 7$ degrees of freedom.

STEP 9

Compare the p-value with $\alpha = 0.05$ .
- If $p \leq \alpha$ , reject the null hypothesis. - If $p > \alpha$ , fail to reject the null hypothesis.

STEP 10

State the conclusion:
Based on the calculated p-value and comparison with $\alpha$ , we can conclude whether there is sufficient evidence to suggest a difference in performance between the two conditions.
The results are statistically significant at $\alpha=0.05$ , so there is sufficient evidence to conclude that the population mean test score taking the exam alone is not the same as the population mean test score taking the exam in a classroom setting.

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Studdy Solution

STEP 1

1. The exam scores for both conditions (alone and classroom) are normally distributed.
2. The sample size is small (n = 8), so a paired t-test is appropriate.
3. The significance level is $\alpha = 0.05$ .

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the differences between paired scores.
3. Compute the test statistic.
4. Determine the p-value.
5. Compare the p-value with $\alpha$ and make a decision.
6. State the conclusion.

STEP 3

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean difference in scores between the two conditions is zero. - Alternative hypothesis ( $H_1$ ): The mean difference in scores between the two conditions is not zero.

STEP 4

Calculate the differences between paired scores:
\begin{align} \text{Differences} & = (81 - 77), (83 - 80), (90 - 78), (89 - 84), (92 - 89), (89 - 86), (82 - 86), (69 - 67) \\ & = 4, 3, 12, 5, 3, 3, -4, 2 \end{align}

STEP 5

Compute the test statistic:
1. Calculate the mean of differences: $\bar{d} = \frac{4 + 3 + 12 + 5 + 3 + 3 - 4 + 2}{8} = 3.5$
2. Calculate the standard deviation of differences: $s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}$
3. Compute the test statistic: $t = \frac{\bar{d}}{s_d / \sqrt{n}}$

STEP 6

Calculate the standard deviation of differences:
$s_d = \sqrt{\frac{(4-3.5)^2 + (3-3.5)^2 + (12-3.5)^2 + (5-3.5)^2 + (3-3.5)^2 + (3-3.5)^2 + (-4-3.5)^2 + (2-3.5)^2}{8-1}}$
$s_d = \sqrt{\frac{0.25 + 0.25 + 72.25 + 2.25 + 0.25 + 0.25 + 56.25 + 2.25}{7}}$
$s_d = \sqrt{\frac{134}{7}} \approx 4.368$

STEP 7

Compute the test statistic:
$t = \frac{3.5}{4.368 / \sqrt{8}} \approx \frac{3.5}{1.544} \approx 2.267$

STEP 8

Determine the p-value using the t-distribution with $n-1 = 7$ degrees of freedom.

STEP 9

Compare the p-value with $\alpha = 0.05$ .
- If $p \leq \alpha$ , reject the null hypothesis. - If $p > \alpha$ , fail to reject the null hypothesis.

STEP 10

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Studdy Solution

STEP 1

1. The exam scores for both conditions (alone and classroom) are normally distributed.2. The sample size is small (n = 8), so a paired t-test is appropriate.3. The significance level is α=0.05\alpha = 0.05α=0.05.

STEP 2

1. State the null and alternative hypotheses.2. Calculate the differences between paired scores.3. Compute the test statistic.4. Determine the p-value.5. Compare the p-value with α\alphaα and make a decision.6. State the conclusion.

STEP 3

State the null and alternative hypotheses:- Null hypothesis (H0H_0H0​): The mean difference in scores between the two conditions is zero. - Alternative hypothesis (H1H_1H1​): The mean difference in scores between the two conditions is not zero.

STEP 4

Calculate the differences between paired scores:\begin{align*} \text{Differences} & = (81 - 77), (83 - 80), (90 - 78), (89 - 84), (92 - 89), (89 - 86), (82 - 86), (69 - 67) \\ & = 4, 3, 12, 5, 3, 3, -4, 2 \end{align*}

STEP 5

STEP 6

STEP 7

Compute the test statistic:t=3.54.368/8≈3.51.544≈2.267t = \frac{3.5}{4.368 / \sqrt{8}} \approx \frac{3.5}{1.544} \approx 2.267t=4.368/8​3.5​≈1.5443.5​≈2.267

STEP 8

Determine the p-value using the t-distribution with n−1=7n-1 = 7n−1=7 degrees of freedom.

STEP 9

Compare the p-value with α=0.05\alpha = 0.05α=0.05.- If p≤αp \leq \alphap≤α, reject the null hypothesis. - If p>αp > \alphap>α, fail to reject the null hypothesis.

STEP 10

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1. The exam scores for both conditions (alone and classroom) are normally distributed.
2. The sample size is small (n = 8), so a paired t-test is appropriate.
3. The significance level is $\alpha = 0.05$ .

1. State the null and alternative hypotheses.
2. Calculate the differences between paired scores.
3. Compute the test statistic.
4. Determine the p-value.
5. Compare the p-value with $\alpha$ and make a decision.
6. State the conclusion.

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean difference in scores between the two conditions is zero. - Alternative hypothesis ( $H_1$ ): The mean difference in scores between the two conditions is not zero.

Calculate the differences between paired scores:
\begin{align} \text{Differences} & = (81 - 77), (83 - 80), (90 - 78), (89 - 84), (92 - 89), (89 - 86), (82 - 86), (69 - 67) \\ & = 4, 3, 12, 5, 3, 3, -4, 2 \end{align}

Compute the test statistic:
$t = \frac{3.5}{4.368 / \sqrt{8}} \approx \frac{3.5}{1.544} \approx 2.267$

Determine the p-value using the t-distribution with $n-1 = 7$ degrees of freedom.

Compare the p-value with $\alpha = 0.05$ .
- If $p \leq \alpha$ , reject the null hypothesis. - If $p > \alpha$ , fail to reject the null hypothesis.