Answered

Get expert advice and community support for your questions on IDNLearn.com. Find in-depth and accurate answers to all your questions from our knowledgeable and dedicated community members.

Using the examscores dataset, a teacher found that the mean score for Exam2 is 79.4, which is below the expected mean of 83. At the [tex]$\alpha=0.05$[/tex] significance level, does sufficient evidence exist that the mean score of the class is lower than the expected mean? Use the output below.

\begin{tabular}{l|r}
\hline
Mean & 79.4 \\
\hline
Variance & 205.4286 \\
\hline
Observations & 50 \\
\hline
Hypothesized Mean & 83 \\
\hline
[tex]$df$[/tex] & 49 \\
\hline
[tex]$t$[/tex] Stat & -1.77606 \\
\hline
[tex]$P(T\ \textless \ =t)$[/tex] one-tail & 0.040966 \\
\hline
[tex]$t$[/tex] Critical one-tail & 1.676551 \\
\hline
[tex]$P(T\ \textless \ =t)$[/tex] two-tail & 0.081933 \\
\hline
[tex]$t$[/tex] Critical two-tail & 2.009575 \\
\hline
\end{tabular}

1. What is the null hypothesis [tex]$H_0$[/tex]?
[tex]\[
\mu = 83
\][/tex]

2. What is the alternative hypothesis [tex]$H_a$[/tex]?
[tex]\[
\mu \ \textless \ 83
\][/tex]

Sagot :

GinnyAnsweridn
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]
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Thanks for visiting IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more helpful information.