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Let's walk through the steps to determine whether there is sufficient evidence to conclude that the mean score of the class is significantly lower than the expected mean at the [tex]\(\alpha = 0.05\)[/tex] significance level.
### Step 1: Formulate Hypotheses
Firstly, we need to set up our hypotheses.
1. Null Hypothesis, [tex]\( H_0 \)[/tex]:
- The null hypothesis is a statement that there is no effect or no difference. In this context, we are testing whether the mean score of the class is equal to the expected mean.
[tex]\[ H_0: \mu = 83 \][/tex]
2. Alternative Hypothesis, [tex]\( H_a \)[/tex]:
- The alternative hypothesis is what we want to test for; it suggests that the mean score of the class is less than the expected mean.
[tex]\[ H_a: \mu < 83 \][/tex]
### Step 2: Evaluate Test Statistics and P-Value
In the provided data:
- The t-statistic is given as [tex]\(-1.77606\)[/tex].
- The one-tail p-value is [tex]\(0.040966\)[/tex].
- The one-tail t-critical value is [tex]\(1.676551\)[/tex].
- The significance level, [tex]\(\alpha\)[/tex], is [tex]\(0.05\)[/tex].
### Step 3: Decision Rule
To make a decision, we compare the p-value to the significance level, [tex]\(\alpha\)[/tex]:
- If the p-value is less than or equal to [tex]\(\alpha\)[/tex], we reject the null hypothesis.
- If the p-value is greater than [tex]\(\alpha\)[/tex], we fail to reject the null hypothesis.
### Step 4: Making a Decision
The provided p-value ([tex]\(0.040966\)[/tex]) is smaller than the significance level ([tex]\(0.05\)[/tex]). Therefore, we have sufficient evidence to reject the null hypothesis.
### Conclusion
Since the p-value is less than [tex]\(\alpha\)[/tex], we reject the null hypothesis. This means that there is sufficient evidence at the [tex]\(\alpha = 0.05\)[/tex] significance level to conclude that the mean score of the class is significantly lower than the expected mean of 83.
### Summary of Hypotheses:
1. Null Hypothesis [tex]\( H_0 \)[/tex]:
[tex]\[ H_0: \mu = 83 \][/tex]
2. Alternative Hypothesis [tex]\( H_a \)[/tex]:
[tex]\[ H_a: \mu < 83 \][/tex]
### Step 1: Formulate Hypotheses
Firstly, we need to set up our hypotheses.
1. Null Hypothesis, [tex]\( H_0 \)[/tex]:
- The null hypothesis is a statement that there is no effect or no difference. In this context, we are testing whether the mean score of the class is equal to the expected mean.
[tex]\[ H_0: \mu = 83 \][/tex]
2. Alternative Hypothesis, [tex]\( H_a \)[/tex]:
- The alternative hypothesis is what we want to test for; it suggests that the mean score of the class is less than the expected mean.
[tex]\[ H_a: \mu < 83 \][/tex]
### Step 2: Evaluate Test Statistics and P-Value
In the provided data:
- The t-statistic is given as [tex]\(-1.77606\)[/tex].
- The one-tail p-value is [tex]\(0.040966\)[/tex].
- The one-tail t-critical value is [tex]\(1.676551\)[/tex].
- The significance level, [tex]\(\alpha\)[/tex], is [tex]\(0.05\)[/tex].
### Step 3: Decision Rule
To make a decision, we compare the p-value to the significance level, [tex]\(\alpha\)[/tex]:
- If the p-value is less than or equal to [tex]\(\alpha\)[/tex], we reject the null hypothesis.
- If the p-value is greater than [tex]\(\alpha\)[/tex], we fail to reject the null hypothesis.
### Step 4: Making a Decision
The provided p-value ([tex]\(0.040966\)[/tex]) is smaller than the significance level ([tex]\(0.05\)[/tex]). Therefore, we have sufficient evidence to reject the null hypothesis.
### Conclusion
Since the p-value is less than [tex]\(\alpha\)[/tex], we reject the null hypothesis. This means that there is sufficient evidence at the [tex]\(\alpha = 0.05\)[/tex] significance level to conclude that the mean score of the class is significantly lower than the expected mean of 83.
### Summary of Hypotheses:
1. Null Hypothesis [tex]\( H_0 \)[/tex]:
[tex]\[ H_0: \mu = 83 \][/tex]
2. Alternative Hypothesis [tex]\( H_a \)[/tex]:
[tex]\[ H_a: \mu < 83 \][/tex]
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