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To solve this hypothesis testing problem, we need to follow several steps.
First, let's summarize the given information:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 78.2\)[/tex]
- Alternative hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu < 78.2\)[/tex] (this is a left-tailed test)
- Significance level ([tex]\(\alpha\)[/tex]): 0.10
- Population standard deviation ([tex]\(\sigma\)[/tex]): 20.9
- Sample mean ([tex]\(\bar{x}\)[/tex]): 75.3
- Sample size ([tex]\(n\)[/tex]): 53
### Step 1: Calculate the Standard Error of the Mean
The standard error of the mean (SE) is calculated using the formula:
[tex]\[ SE = \frac{\sigma}{\sqrt{n}} \][/tex]
Substituting the known values:
[tex]\[ SE = \frac{20.9}{\sqrt{53}} \][/tex]
### Step 2: Calculate the Test Statistic (z-score)
The test statistic (z-score) for a single sample mean is calculated using the formula:
[tex]\[ z = \frac{\bar{x} - \mu}{SE} \][/tex]
Using the values:
[tex]\[ z = \frac{75.3 - 78.2}{SE} \][/tex]
### Step 3: Calculate the p-value
The p-value for a left-tailed test corresponds to the area under the standard normal curve to the left of the calculated z-score. This can be found using statistical tables or software tools.
### Step 4: Compare the p-value with [tex]\(\alpha\)[/tex] and make a decision
If the p-value is less than or equal to the significance level ([tex]\(\alpha = 0.10\)[/tex]), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
### Solution:
- The standard error is:
[tex]\[ SE = \frac{20.9}{\sqrt{53}} \approx 2.87 \][/tex]
- The test statistic (z-score) is:
[tex]\[ z = \frac{75.3 - 78.2}{2.87} \approx -1.01 \][/tex]
- The p-value for [tex]\( z = -1.01 \)[/tex] is approximately [tex]\( 0.1562 \)[/tex].
Comparing the p-value to the significance level:
- Since [tex]\( 0.1562 \)[/tex] is greater than [tex]\( \alpha = 0.10 \)[/tex], we fail to reject the null hypothesis.
### Final Answers:
1. Test Statistic [tex]\( z \)[/tex] = [tex]\(-1.01\)[/tex] (accurate to 3 decimal places)
2. p-value = [tex]\( 0.1562 \)[/tex] (accurate to 4 decimal places)
3. The p-value is greater than [tex]\(\alpha\)[/tex].
4. The test statistic leads to a decision to fail to reject the null hypothesis.
First, let's summarize the given information:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 78.2\)[/tex]
- Alternative hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu < 78.2\)[/tex] (this is a left-tailed test)
- Significance level ([tex]\(\alpha\)[/tex]): 0.10
- Population standard deviation ([tex]\(\sigma\)[/tex]): 20.9
- Sample mean ([tex]\(\bar{x}\)[/tex]): 75.3
- Sample size ([tex]\(n\)[/tex]): 53
### Step 1: Calculate the Standard Error of the Mean
The standard error of the mean (SE) is calculated using the formula:
[tex]\[ SE = \frac{\sigma}{\sqrt{n}} \][/tex]
Substituting the known values:
[tex]\[ SE = \frac{20.9}{\sqrt{53}} \][/tex]
### Step 2: Calculate the Test Statistic (z-score)
The test statistic (z-score) for a single sample mean is calculated using the formula:
[tex]\[ z = \frac{\bar{x} - \mu}{SE} \][/tex]
Using the values:
[tex]\[ z = \frac{75.3 - 78.2}{SE} \][/tex]
### Step 3: Calculate the p-value
The p-value for a left-tailed test corresponds to the area under the standard normal curve to the left of the calculated z-score. This can be found using statistical tables or software tools.
### Step 4: Compare the p-value with [tex]\(\alpha\)[/tex] and make a decision
If the p-value is less than or equal to the significance level ([tex]\(\alpha = 0.10\)[/tex]), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
### Solution:
- The standard error is:
[tex]\[ SE = \frac{20.9}{\sqrt{53}} \approx 2.87 \][/tex]
- The test statistic (z-score) is:
[tex]\[ z = \frac{75.3 - 78.2}{2.87} \approx -1.01 \][/tex]
- The p-value for [tex]\( z = -1.01 \)[/tex] is approximately [tex]\( 0.1562 \)[/tex].
Comparing the p-value to the significance level:
- Since [tex]\( 0.1562 \)[/tex] is greater than [tex]\( \alpha = 0.10 \)[/tex], we fail to reject the null hypothesis.
### Final Answers:
1. Test Statistic [tex]\( z \)[/tex] = [tex]\(-1.01\)[/tex] (accurate to 3 decimal places)
2. p-value = [tex]\( 0.1562 \)[/tex] (accurate to 4 decimal places)
3. The p-value is greater than [tex]\(\alpha\)[/tex].
4. The test statistic leads to a decision to fail to reject the null hypothesis.
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