Question 1165658: A random sample of 30 students was selected from a large group of students taking a certain calculus test. The mean score for the students in the sample was 88. Assume that σ = 7.4. Construct a 99% confidence interval for the mean score, μ, of all students taking the test.
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website! You can construct the $99\%$ confidence interval using the formula for a population mean when the population standard deviation ($\sigma$) is known (the $Z$-interval).
The resulting $99\%$ confidence interval for the mean score ($\mu$) of all students is **$84.52$ to $91.48$**.
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## 1. Identify the Key Values
| Symbol | Description | Value |
| :---: | :--- | :--- |
| $\bar{x}$ | Sample Mean | 88 |
| $\sigma$ | Population Standard Deviation | 7.4 |
| $n$ | Sample Size | 30 |
| C.L. | Confidence Level | 99% |
| $Z_{\alpha/2}$ | Critical Z-score for 99% CI | 2.576 |
### Finding the Critical Z-score
For a $99\%$ confidence level, the area in the two tails is $1 - 0.99 = 0.01$, and the area in one tail ($\alpha/2$) is $0.005$. The corresponding $Z$-score is $\mathbf{2.576}$.
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## 2. Calculate the Margin of Error ($ME$)
The margin of error is calculated as:
$$ME = Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$$
$$ME = 2.576 \cdot \frac{7.4}{\sqrt{30}}$$
$$ME \approx 2.576 \cdot \frac{7.4}{5.4772}$$
$$ME \approx 2.576 \cdot 1.3510$$
$$\mathbf{ME \approx 3.48}$$
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## 3. Construct the Confidence Interval
The confidence interval is calculated as:
$$\text{Confidence Interval} = \bar{x} \pm ME$$
$$\text{C.I.} = 88 \pm 3.48$$
$$\text{Lower Bound} = 88 - 3.48 = \mathbf{84.52}$$
$$\text{Upper Bound} = 88 + 3.48 = \mathbf{91.48}$$
The $99\%$ confidence interval for the mean score of all students is **$(84.52, 91.48)$**.
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