Math Problem Statement

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.)

Solution

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 p^\hat{p} is given by the formula:

p^=xn\hat{p} = \frac{x}{n}

Where:

  • x=142x = 142 is the number of distorted graphics.
  • n=453n = 453 is the total number of graphics.

Now, calculate p^\hat{p}:

p^=1424530.3135\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:

p^±Zα/2p^(1p^)n\hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}

Where:

  • p^\hat{p} is the point estimate of the proportion (0.3135).
  • Zα/2Z_{\alpha/2} is the critical value for the 95% confidence level (which is 1.96).
  • nn is the sample size (453).

Now, calculate the margin of error:

Margin of error=1.96×0.3135(10.3135)453\text{Margin of error} = 1.96 \times \sqrt{\frac{0.3135(1 - 0.3135)}{453}}

=1.96×0.3135×0.6865453= 1.96 \times \sqrt{\frac{0.3135 \times 0.6865}{453}}

=1.96×0.2151453= 1.96 \times \sqrt{\frac{0.2151}{453}}

=1.96×0.0004748= 1.96 \times \sqrt{0.0004748}

=1.96×0.021790.0427= 1.96 \times 0.02179 \approx 0.0427

Now, calculate the confidence interval:

p^±0.0427\hat{p} \pm 0.0427

0.31350.0427=0.27080.3135 - 0.0427 = 0.2708 0.3135+0.0427=0.35620.3135 + 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 Zα/2Z_{\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!

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Proportions

Formulas

Point estimate formula: \hat{p} = x / n
Confidence interval for proportions: \hat{p} ± Z_{\alpha/2} * sqrt(\hat{p}(1 - \hat{p}) / n)

Theorems

Central Limit Theorem
Normal Distribution Properties

Suitable Grade Level

College Level Statistics