Question

Sample 124, 35 of them score Ed less than 530.
Find a point estimate for those who scored less than 530

Answer 1

Construction a 99% confidence

Answer 2

To find the point estimate for the proportion of individuals in the sample who scored less than 530, you can use the formula for the sample proportion:

\[
\hat{p} = \frac{x}{n}
\]

where:
- \( \hat{p} \) is the sample proportion,
- \( x \) is the number of successes (in this case, the number of individuals who scored less than 530),
- \( n \) is the total sample size.

In your case:
- \( x = 35 \) (the number of individuals who scored less than 530),
- \( n = 124 \) (the total number of individuals in the sample).

Now, plug in the values:

\[
\hat{p} = \frac{35}{124} \approx 0.2823
\]

Thus, the point estimate for the proportion of the population that scored less than 530 is approximately **0.2823** or **28.23%**.

Answer 3

To construct a 99% confidence interval for the proportion of individuals who scored less than 530 based on the provided sample data, we will follow these steps:

1. **Calculate the sample proportion** \((\hat{p})\).
2. **Calculate the standard error** (SE) of the sample proportion.
3. **Find the Z-score** for the desired confidence level.
4. **Construct the confidence interval** using the sample proportion, the standard error, and the Z-score.

### Step 1: Calculate the Sample Proportion

We calculated the sample proportion in the previous response:

\[
\hat{p} = \frac{35}{124} \approx 0.2823
\]

### Step 2: Calculate the Standard Error (SE)

The standard error for the sample proportion is given by the formula:

\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]

Plugging in the values:

\[
SE = \sqrt{\frac{0.2823(1 - 0.2823)}{124}} = \sqrt{\frac{0.2823 \times 0.7177}{124}} \approx \sqrt{\frac{0.2028}{124}} \approx \sqrt{0.001634} \approx 0.0404
\]

### Step 3: Find the Z-Score for 99% Confidence Level

For a 99% confidence level, the Z-score (critical value) can be found from the standard normal distribution. The Z-score is approximately \(2.576\) for a 99% confidence level.

### Step 4: Construct the Confidence Interval

The confidence interval is calculated using the formula:

\[
\hat{p} \pm Z \cdot SE
\]

Substituting the values:

\[
0.2823 \pm 2.576 \cdot 0.0404
\]

Calculating the margin of error:

\[
2.576 \cdot 0.0404 \approx 0.1041
\]

Now we can construct the confidence interval:

\[
(0.2823 - 0.1041, 0.2823 + 0.1041)
\]

Calculating the interval:

\[
(0.1782, 0.3864)
\]

### Conclusion

The 99% confidence interval for the proportion of individuals who scored less than 530 is approximately **(0.1782, 0.3864)**, or **(17.82%, 38.64%)**. This means you can be 99% confident that the true proportion of individuals in the population who scored less than 530 lies within this interval.