Question
What is the probability of rejecting the null hypothesis in a two-tailed test when the population mean is 50, the sample mean is 55, the sample size is 100, and the population standard deviation is 10?
Answers
GPT-4o mini
To find the probability of rejecting the null hypothesis in a two-tailed test, we need to calculate the test statistic and then determine the rejection region based on the chosen significance level (alpha). However, the question does not specify the significance level, so we will go ahead and simplify the calculations assuming an alpha level of 0.05 for this example.
### Step 1: Define the Null and Alternative Hypotheses
- Null hypothesis \( H_0: \mu = 50 \)
- Alternative hypothesis \( H_a: \mu \neq 50 \)
### Step 2: Calculate the Test Statistic
The formula for the z-test statistic for a sample mean is:
\[
z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}
\]
where:
- \( \bar{x} \) = sample mean = 55
- \( \mu_0 \) = population mean under the null hypothesis = 50
- \( \sigma \) = population standard deviation = 10
- \( n \) = sample size = 100
Substituting the values into the formula:
\[
z = \frac{55 - 50}{10 / \sqrt{100}} = \frac{5}{10 / 10} = \frac{5}{1} = 5
\]
### Step 3: Find Critical Values for Two-Tailed Test
For a significance level of \( \alpha = 0.05 \), the critical z-values for a two-tailed test are approximately:
- Critical z-value for lower tail: \( -1.96 \)
- Critical z-value for upper tail: \( 1.96 \)
### Step 4: Calculate the p-value
To find the p-value associated with our calculated z-value of 5:
1. The z-score is significantly greater than the critical z-value of 1.96.
2. Using standard normal distribution tables or a calculator:
The area to the right of z = 5 (p-value) is very small, close to zero.
Since this is a two-tailed test, we multiply the p-value by 2.
### Step 5: Conclusion
If the p-value (which is very small, essentially 0) is less than the alpha level (0.05), we reject the null hypothesis. In this case, the p-value is so low that it leads us to reject the null hypothesis.
Since we are looking for the probability of rejecting the null hypothesis given that \( \mu = 50 \):
In conclusion, if we compute the probability of rejecting the null hypothesis given the sample mean of 55, we find that it will be nearly 1 (or 100%) since our z-value is significantly beyond the critical value. Thus, the probability of rejecting the null hypothesis in this context is essentially 1.
### Step 1: Define the Null and Alternative Hypotheses
- Null hypothesis \( H_0: \mu = 50 \)
- Alternative hypothesis \( H_a: \mu \neq 50 \)
### Step 2: Calculate the Test Statistic
The formula for the z-test statistic for a sample mean is:
\[
z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}
\]
where:
- \( \bar{x} \) = sample mean = 55
- \( \mu_0 \) = population mean under the null hypothesis = 50
- \( \sigma \) = population standard deviation = 10
- \( n \) = sample size = 100
Substituting the values into the formula:
\[
z = \frac{55 - 50}{10 / \sqrt{100}} = \frac{5}{10 / 10} = \frac{5}{1} = 5
\]
### Step 3: Find Critical Values for Two-Tailed Test
For a significance level of \( \alpha = 0.05 \), the critical z-values for a two-tailed test are approximately:
- Critical z-value for lower tail: \( -1.96 \)
- Critical z-value for upper tail: \( 1.96 \)
### Step 4: Calculate the p-value
To find the p-value associated with our calculated z-value of 5:
1. The z-score is significantly greater than the critical z-value of 1.96.
2. Using standard normal distribution tables or a calculator:
The area to the right of z = 5 (p-value) is very small, close to zero.
Since this is a two-tailed test, we multiply the p-value by 2.
### Step 5: Conclusion
If the p-value (which is very small, essentially 0) is less than the alpha level (0.05), we reject the null hypothesis. In this case, the p-value is so low that it leads us to reject the null hypothesis.
Since we are looking for the probability of rejecting the null hypothesis given that \( \mu = 50 \):
In conclusion, if we compute the probability of rejecting the null hypothesis given the sample mean of 55, we find that it will be nearly 1 (or 100%) since our z-value is significantly beyond the critical value. Thus, the probability of rejecting the null hypothesis in this context is essentially 1.