Question
Sample 10 people took a test. Can you conclidebthebmean increase in scores differed from 15 points? Use a=.10
Before- 408
378
466
470
473
443
459
462
493
382
After
407
396
488
489
473
448
473
428
525
382
State the null and alternative hypothesis
Before- 408
378
466
470
473
443
459
462
493
382
After
407
396
488
489
473
448
473
428
525
382
State the null and alternative hypothesis
Answers
GPT-4o mini
To test whether the mean increase in scores differed from 15 points, we will set up our null and alternative hypotheses.
### Null Hypothesis (\(H_0\)):
The mean increase in scores is equal to 15 points. Mathematically, this can be represented as:
\[
H_0: \mu = 15
\]
### Alternative Hypothesis (\(H_a\)):
The mean increase in scores is not equal to 15 points. This can be represented as:
\[
H_a: \mu \neq 15
\]
### Steps for Testing:
1. Calculate the differences in scores for each subject (after - before).
2. Calculate the mean of these differences.
3. Perform a t-test (since the sample size is small and we may not know the population standard deviation) to test the hypotheses at the alpha level of 0.10.
Once you calculate the differences and conduct the t-test, you can draw conclusions about the null and alternative hypotheses based on the calculated p-value and critical values.
### Null Hypothesis (\(H_0\)):
The mean increase in scores is equal to 15 points. Mathematically, this can be represented as:
\[
H_0: \mu = 15
\]
### Alternative Hypothesis (\(H_a\)):
The mean increase in scores is not equal to 15 points. This can be represented as:
\[
H_a: \mu \neq 15
\]
### Steps for Testing:
1. Calculate the differences in scores for each subject (after - before).
2. Calculate the mean of these differences.
3. Perform a t-test (since the sample size is small and we may not know the population standard deviation) to test the hypotheses at the alpha level of 0.10.
Once you calculate the differences and conduct the t-test, you can draw conclusions about the null and alternative hypotheses based on the calculated p-value and critical values.
Computer the p value round to 4 places
GPT-4o mini
To compute the p-value for the given data, follow these steps:
1. **Calculate the differences between the "After" scores and "Before" scores:**
| Before | After | Difference (After - Before) |
|--------|-------|-----------------------------|
| 408 | 407 | -1 |
| 378 | 396 | 18 |
| 466 | 488 | 22 |
| 470 | 489 | 19 |
| 473 | 473 | 0 |
| 443 | 448 | 5 |
| 459 | 473 | 14 |
| 462 | 428 | -34 |
| 493 | 525 | 32 |
| 382 | 382 | 0 |
Now, calculate the differences:
\[
\text{Differences} = [-1, 18, 22, 19, 0, 5, 14, -34, 32, 0]
\]
2. **Calculate the mean of the differences:**
\[
\text{Mean} = \frac{\sum (\text{Differences})}{n}
\]
\[
\sum (\text{Differences}) = -1 + 18 + 22 + 19 + 0 + 5 + 14 - 34 + 32 + 0 = 75
\]
\[
n = 10
\]
\[
\text{Mean} = \frac{75}{10} = 7.5
\]
3. **Calculate the standard deviation of the differences:**
\[
\text{SD} = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}}
\]
where \(d_i\) is each difference, and \(\bar{d}\) is the mean of the differences.
First, calculate the squared differences:
\[
d_i - \bar{d} = [-8.5, 10.5, 14.5, 11.5, -7.5, -2.5, 6.5, -41.5, 24.5, -7.5]
\]
Now square these differences:
\[
[72.25, 110.25, 210.25, 132.25, 56.25, 6.25, 42.25, 1722.25, 600.25, 56.25]
\]
Sum the squared differences:
\[
\sum (d_i - \bar{d})^2 = 72.25 + 110.25 + 210.25 + 132.25 + 56.25 + 6.25 + 42.25 + 1722.25 + 600.25 + 56.25 = 2119.5
\]
Now calculate the standard deviation:
\[
\text{SD} = \sqrt{\frac{2119.5}{10 - 1}} = \sqrt{\frac{2119.5}{9}} \approx 15.36
\]
4. **Calculate the t-statistic:**
Using the t-statistic formula for one sample:
\[
t = \frac{\bar{d} - \mu_0}{\frac{\text{SD}}{\sqrt{n}}}
\]
where \(\mu_0 = 15\).
\[
t = \frac{7.5 - 15}{\frac{15.36}{\sqrt{10}}} = \frac{-7.5}{\frac{15.36}{3.162}} \approx \frac{-7.5}{4.86} \approx -1.54
\]
5. **Find the p-value:**
The degrees of freedom (df) = n - 1 = 9.
Using a t-distribution table (or a calculator), we can find the two-tailed p-value for \(t \approx -1.54\) with 9 degrees of freedom.
Looking this up or using a calculator, we find the p-value to be approximately:
\[
p \approx 0.1573
\]
### Final Result:
Thus, the p-value rounded to four decimal places is:
\[
\text{p-value} \approx 0.1573
\]
1. **Calculate the differences between the "After" scores and "Before" scores:**
| Before | After | Difference (After - Before) |
|--------|-------|-----------------------------|
| 408 | 407 | -1 |
| 378 | 396 | 18 |
| 466 | 488 | 22 |
| 470 | 489 | 19 |
| 473 | 473 | 0 |
| 443 | 448 | 5 |
| 459 | 473 | 14 |
| 462 | 428 | -34 |
| 493 | 525 | 32 |
| 382 | 382 | 0 |
Now, calculate the differences:
\[
\text{Differences} = [-1, 18, 22, 19, 0, 5, 14, -34, 32, 0]
\]
2. **Calculate the mean of the differences:**
\[
\text{Mean} = \frac{\sum (\text{Differences})}{n}
\]
\[
\sum (\text{Differences}) = -1 + 18 + 22 + 19 + 0 + 5 + 14 - 34 + 32 + 0 = 75
\]
\[
n = 10
\]
\[
\text{Mean} = \frac{75}{10} = 7.5
\]
3. **Calculate the standard deviation of the differences:**
\[
\text{SD} = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}}
\]
where \(d_i\) is each difference, and \(\bar{d}\) is the mean of the differences.
First, calculate the squared differences:
\[
d_i - \bar{d} = [-8.5, 10.5, 14.5, 11.5, -7.5, -2.5, 6.5, -41.5, 24.5, -7.5]
\]
Now square these differences:
\[
[72.25, 110.25, 210.25, 132.25, 56.25, 6.25, 42.25, 1722.25, 600.25, 56.25]
\]
Sum the squared differences:
\[
\sum (d_i - \bar{d})^2 = 72.25 + 110.25 + 210.25 + 132.25 + 56.25 + 6.25 + 42.25 + 1722.25 + 600.25 + 56.25 = 2119.5
\]
Now calculate the standard deviation:
\[
\text{SD} = \sqrt{\frac{2119.5}{10 - 1}} = \sqrt{\frac{2119.5}{9}} \approx 15.36
\]
4. **Calculate the t-statistic:**
Using the t-statistic formula for one sample:
\[
t = \frac{\bar{d} - \mu_0}{\frac{\text{SD}}{\sqrt{n}}}
\]
where \(\mu_0 = 15\).
\[
t = \frac{7.5 - 15}{\frac{15.36}{\sqrt{10}}} = \frac{-7.5}{\frac{15.36}{3.162}} \approx \frac{-7.5}{4.86} \approx -1.54
\]
5. **Find the p-value:**
The degrees of freedom (df) = n - 1 = 9.
Using a t-distribution table (or a calculator), we can find the two-tailed p-value for \(t \approx -1.54\) with 9 degrees of freedom.
Looking this up or using a calculator, we find the p-value to be approximately:
\[
p \approx 0.1573
\]
### Final Result:
Thus, the p-value rounded to four decimal places is:
\[
\text{p-value} \approx 0.1573
\]
Do we reject Hv0
GPT-4o mini
To determine whether to reject the null hypothesis (\(H_0\)) based on the calculated p-value, we compare the p-value to the significance level \(\alpha\) that you specified.
### Given:
- **Significance level (\(\alpha\)):** 0.10
- **Calculated p-value:** 0.1573
### Decision Rule:
- If the p-value is less than \(\alpha\) (p-value < 0.10), we reject the null hypothesis.
- If the p-value is greater than or equal to \(\alpha\) (p-value ≥ 0.10), we fail to reject the null hypothesis.
### Conclusion:
In this case:
- \(0.1573 > 0.10\)
Since the p-value (0.1573) is greater than the significance level (0.10), **we fail to reject the null hypothesis (\(H_0\))**.
### Interpretation:
This means that there is not enough evidence at the 0.10 significance level to conclude that the mean increase in scores differs from 15 points.
### Given:
- **Significance level (\(\alpha\)):** 0.10
- **Calculated p-value:** 0.1573
### Decision Rule:
- If the p-value is less than \(\alpha\) (p-value < 0.10), we reject the null hypothesis.
- If the p-value is greater than or equal to \(\alpha\) (p-value ≥ 0.10), we fail to reject the null hypothesis.
### Conclusion:
In this case:
- \(0.1573 > 0.10\)
Since the p-value (0.1573) is greater than the significance level (0.10), **we fail to reject the null hypothesis (\(H_0\))**.
### Interpretation:
This means that there is not enough evidence at the 0.10 significance level to conclude that the mean increase in scores differs from 15 points.