Question
Based on a sample of 40 people, the sample mean GPA was 3.18 with a standard deviation of 0.06
The test statistic is: ____
(to 2 decimals)
The p-value is: ____
(to 2 decimals)
The test statistic is: ____
(to 2 decimals)
The p-value is: ____
(to 2 decimals)
Answers
Answered by
GPT-4o mini
To calculate the test statistic for the sample mean GPA, we use the formula for the z-test statistic because the sample size is sufficiently large (n = 40). The formula for the test statistic \( z \) is given by:
\[
z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}
\]
Where:
- \( \bar{x} \) is the sample mean (3.18)
- \( \mu \) is the population mean (we'll need a value for this to compare against; typically, this is known or hypothesized)
- \( s \) is the sample standard deviation (0.06)
- \( n \) is the sample size (40)
Since the population mean \( \mu \) is not provided, I cannot calculate the z score without this value. However, I can guide you through the steps assuming you have a specific \( \mu \) in mind.
Assuming \( \mu \) is provided, you can plug it into the formula and calculate the z-score.
### Example Calculation
Let's say you want to test against a hypothesized population mean \( \mu = 3.0 \).
1. Calculate the standard error (SE):
\[
SE = \frac{s}{\sqrt{n}} = \frac{0.06}{\sqrt{40}} \approx \frac{0.06}{6.32} \approx 0.00946
\]
2. Calculate the z-score:
\[
z = \frac{3.18 - 3.00}{0.00946} \approx \frac{0.18}{0.00946} \approx 19.03
\]
Now let's find the p-value corresponding to this z-score. In this case, we will need to refer to the standard normal distribution table (or use a calculator).
Assuming a two-tailed test for the hypothesis, a z-score of approximately 19.03 would lead to an extremely small p-value, close to 0.
### Summary (with hypothetical \( \mu = 3.0 \)):
- Test statistic \( z \): 19.03 (2 decimal places would be 19.03)
- p-value: Essentially 0 (as it's way below 0.01)
If you have a different hypothesized population mean, please provide it so that I can calculate the exact test statistic and p-value.
\[
z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}
\]
Where:
- \( \bar{x} \) is the sample mean (3.18)
- \( \mu \) is the population mean (we'll need a value for this to compare against; typically, this is known or hypothesized)
- \( s \) is the sample standard deviation (0.06)
- \( n \) is the sample size (40)
Since the population mean \( \mu \) is not provided, I cannot calculate the z score without this value. However, I can guide you through the steps assuming you have a specific \( \mu \) in mind.
Assuming \( \mu \) is provided, you can plug it into the formula and calculate the z-score.
### Example Calculation
Let's say you want to test against a hypothesized population mean \( \mu = 3.0 \).
1. Calculate the standard error (SE):
\[
SE = \frac{s}{\sqrt{n}} = \frac{0.06}{\sqrt{40}} \approx \frac{0.06}{6.32} \approx 0.00946
\]
2. Calculate the z-score:
\[
z = \frac{3.18 - 3.00}{0.00946} \approx \frac{0.18}{0.00946} \approx 19.03
\]
Now let's find the p-value corresponding to this z-score. In this case, we will need to refer to the standard normal distribution table (or use a calculator).
Assuming a two-tailed test for the hypothesis, a z-score of approximately 19.03 would lead to an extremely small p-value, close to 0.
### Summary (with hypothetical \( \mu = 3.0 \)):
- Test statistic \( z \): 19.03 (2 decimal places would be 19.03)
- p-value: Essentially 0 (as it's way below 0.01)
If you have a different hypothesized population mean, please provide it so that I can calculate the exact test statistic and p-value.