In an article in Accounting and Business Research, Beattie and Jones investigate the use and abuse of graphic presentations in the annual reports of United Kingdom firms. The authors found that 65 percent of the sampled companies graph at least one key financial variable, but that 30 percent of the graphics are materially distorted (nonzero vertical axis, exaggerated trend, or the like). Results for U.S. firms have been found to be similar.

(a) Suppose that in a random sample of 453 graphics from the annual reports of United Kingdom firms, 142 of the graphics are found to be distorted. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.K. annual report graphics that are distorted. (Round your answers to 4 decimal places.)

To solve this problem, we will calculate two things:

1. The **point estimate** for the proportion of distorted graphics.
2. The **95% confidence interval** for the proportion of distorted graphics.

### (1) **Point Estimate**

The point estimate for the proportion \( \hat{p} \) is given by the formula:

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

Where:
- \( x = 142 \) is the number of distorted graphics.
- \( n = 453 \) is the total number of graphics.

Now, calculate \( \hat{p} \):

\[
\hat{p} = \frac{142}{453} \approx 0.3135
\]

So, the point estimate for the proportion of distorted graphics is **0.3135**.

### (2) **95% Confidence Interval**

The formula for the confidence interval for a proportion is:

\[
\hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]

Where:
- \( \hat{p} \) is the point estimate of the proportion (0.3135).
- \( Z_{\alpha/2} \) is the critical value for the 95% confidence level (which is 1.96).
- \( n \) is the sample size (453).

Now, calculate the margin of error:

\[
\text{Margin of error} = 1.96 \times \sqrt{\frac{0.3135(1 - 0.3135)}{453}} 
\]

\[
= 1.96 \times \sqrt{\frac{0.3135 \times 0.6865}{453}} 
\]

\[
= 1.96 \times \sqrt{\frac{0.2151}{453}} 
\]

\[
= 1.96 \times \sqrt{0.0004748} 
\]

\[
= 1.96 \times 0.02179 \approx 0.0427
\]

Now, calculate the confidence interval:

\[
\hat{p} \pm 0.0427
\]

\[
0.3135 - 0.0427 = 0.2708
\]
\[
0.3135 + 0.0427 = 0.3562
\]

So, the 95% confidence interval is approximately **(0.2708, 0.3562)**.

### Final Answer:
- The point estimate of the proportion of distorted graphics is **0.3135**.
- The 95% confidence interval is **(0.2708, 0.3562)**.

Would you like more details or have any further questions? Here are some related questions for deeper understanding:

1. How is the margin of error in a confidence interval influenced by sample size?
2. Why is the \( Z_{\alpha/2} \) value 1.96 for a 95% confidence interval?
3. What does it mean if a confidence interval does not contain a certain value, like 0.30?
4. How does the confidence level (95% in this case) affect the width of the confidence interval?
5. Can the point estimate of a proportion be used to make predictions about a future sample?

**Tip**: A larger sample size will decrease the margin of error, leading to a more precise confidence interval!

Calculate the point estimate and 95% confidence interval for the proportion of distorted graphics in a sample of 453 graphics from U.K. annual reports. The point estimate is 0.3135, with a confidence interval of (0.2708, 0.3562). Learn how to use statistical formulas for proportions.

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