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Answer :
To solve this problem, we need to conduct a hypothesis test for the mean of a normally distributed population when the population standard deviation is unknown. Here's the step-by-step solution:
### Step 1: State the Hypotheses
The null hypothesis ([tex]\(H_0\)[/tex]) and the alternative hypothesis ([tex]\(H_a\)[/tex]) are given as:
- [tex]\(H_0: \mu = 83.8\)[/tex]
- [tex]\(H_a: \mu \neq 83.8\)[/tex]
### Step 2: Collect the Data
We have a sample of data with 55 observations.
### Step 3: Calculate the Sample Statistics
- Sample Mean ([tex]\(\bar{x}\)[/tex]): This is the average of the sample data.
- Sample Standard Deviation (s): This measures the amount of variation or dispersion in the sample data.
- Sample Size (n): The number of data points, which is 55.
### Step 4: Calculate the Test Statistic
The test statistic for a t-test when the population standard deviation is unknown is calculated using the formula:
[tex]\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\][/tex]
Where:
- [tex]\(\bar{x}\)[/tex] is the sample mean,
- [tex]\(\mu_0\)[/tex] is the population mean under the null hypothesis,
- [tex]\(s\)[/tex] is the sample standard deviation,
- [tex]\(n\)[/tex] is the sample size.
### Step 5: Calculate the p-value
Since this is a two-tailed test (because of [tex]\(\mu \neq 83.8\)[/tex] in [tex]\(H_a\)[/tex]), we need to find the probability that the t-statistic is more extreme than the observed value in either direction.
The p-value is calculated as:
[tex]\[
\text{p-value} = 2 \times P(T > |t|)
\][/tex]
Where [tex]\(T\)[/tex] follows a t-distribution with [tex]\(n-1\)[/tex] degrees of freedom.
### Step 6: Compare the p-value with the Significance Level
- The significance level ([tex]\(\alpha\)[/tex]) is 0.01.
- If the p-value is less than [tex]\(\alpha\)[/tex], we reject the null hypothesis.
### Results
Upon calculating, we have:
- Test Statistic [tex]\(t\)[/tex]: [tex]\(-3.814\)[/tex]
- p-value: [tex]\(0.0004\)[/tex]
### Conclusion
Since the p-value [tex]\(0.0004\)[/tex] is less than the significance level [tex]\(0.01\)[/tex], we reject the null hypothesis. There is sufficient evidence to conclude that the mean is significantly different from 83.8.
### Step 1: State the Hypotheses
The null hypothesis ([tex]\(H_0\)[/tex]) and the alternative hypothesis ([tex]\(H_a\)[/tex]) are given as:
- [tex]\(H_0: \mu = 83.8\)[/tex]
- [tex]\(H_a: \mu \neq 83.8\)[/tex]
### Step 2: Collect the Data
We have a sample of data with 55 observations.
### Step 3: Calculate the Sample Statistics
- Sample Mean ([tex]\(\bar{x}\)[/tex]): This is the average of the sample data.
- Sample Standard Deviation (s): This measures the amount of variation or dispersion in the sample data.
- Sample Size (n): The number of data points, which is 55.
### Step 4: Calculate the Test Statistic
The test statistic for a t-test when the population standard deviation is unknown is calculated using the formula:
[tex]\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\][/tex]
Where:
- [tex]\(\bar{x}\)[/tex] is the sample mean,
- [tex]\(\mu_0\)[/tex] is the population mean under the null hypothesis,
- [tex]\(s\)[/tex] is the sample standard deviation,
- [tex]\(n\)[/tex] is the sample size.
### Step 5: Calculate the p-value
Since this is a two-tailed test (because of [tex]\(\mu \neq 83.8\)[/tex] in [tex]\(H_a\)[/tex]), we need to find the probability that the t-statistic is more extreme than the observed value in either direction.
The p-value is calculated as:
[tex]\[
\text{p-value} = 2 \times P(T > |t|)
\][/tex]
Where [tex]\(T\)[/tex] follows a t-distribution with [tex]\(n-1\)[/tex] degrees of freedom.
### Step 6: Compare the p-value with the Significance Level
- The significance level ([tex]\(\alpha\)[/tex]) is 0.01.
- If the p-value is less than [tex]\(\alpha\)[/tex], we reject the null hypothesis.
### Results
Upon calculating, we have:
- Test Statistic [tex]\(t\)[/tex]: [tex]\(-3.814\)[/tex]
- p-value: [tex]\(0.0004\)[/tex]
### Conclusion
Since the p-value [tex]\(0.0004\)[/tex] is less than the significance level [tex]\(0.01\)[/tex], we reject the null hypothesis. There is sufficient evidence to conclude that the mean is significantly different from 83.8.
Thank you for reading the article You wish to test the following claim tex H a tex at a significance level of tex alpha 0 01 tex begin array l H. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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