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.
