Math Problem Statement

The website of the large corporation CorpPlus claims that 93% of the population of all employees say the corporation is a great place to work. You work for a competitor and think that this percentage is not a true indicator of CorpPlus's employee satisfaction. To test the claim from the website, you will survey a random sample of 55 employees from CorpPlus, and ask them whether they think the corporation is a great place to work. Follow the steps below to construct a 90% confidence interval for the population proportion of all employees who say the corporation is a great place to work. Then state whether the confidence interval you construct contradicts the website's claim.

Solution

Let's go through this problem step-by-step.

Step-by-Step Solution

Step 1: Determine the values

From the table, we have:

  • Sample size (n) = 55
  • Number who say "Yes" = 44, so sample proportion (p-hat) = 4455=0.8\frac{44}{55} = 0.8

Step 2: Point Estimate

The point estimate for the sample proportion is p=0.8p = 0.8.

Step 3: Choose the Critical Value for a 90% Confidence Interval

For a 90% confidence interval, the z-score (critical value) is 1.645.

Step 4: Calculate the Standard Error

The formula for the standard error of a proportion is: Standard Error=p(1p)n\text{Standard Error} = \sqrt{\frac{p \cdot (1 - p)}{n}} Plugging in the values: Standard Error=0.8(10.8)55=0.80.255=0.16550.0537\text{Standard Error} = \sqrt{\frac{0.8 \cdot (1 - 0.8)}{55}} = \sqrt{\frac{0.8 \cdot 0.2}{55}} = \sqrt{\frac{0.16}{55}} \approx 0.0537

Step 5: Calculate the Margin of Error

The margin of error (ME) is calculated as: Margin of Error=z×Standard Error=1.645×0.05370.0883\text{Margin of Error} = z \times \text{Standard Error} = 1.645 \times 0.0537 \approx 0.0883

Step 6: Calculate the 90% Confidence Interval

The confidence interval is: Confidence Interval=p±Margin of Error=0.8±0.0883\text{Confidence Interval} = p \pm \text{Margin of Error} = 0.8 \pm 0.0883 So the interval is approximately: (0.7117,0.8883)(0.7117, 0.8883)

Step 7: Answer Part (c)

The claim from the website is that 93% (or 0.93) of employees think the corporation is a great place to work. Since 0.93 is outside the 90% confidence interval of (0.7117, 0.8883), this contradicts the website’s claim.

So, the correct answer to part (c) would be:

  • Yes, the confidence interval contradicts the claim. The proportion 0.93 from the website is outside the 90% confidence interval.

Would you like further details on any part of this solution, or do you have any other questions?

Related Questions

  1. How would the confidence interval change if we used a 95% confidence level instead of 90%?
  2. What would happen to the confidence interval if the sample size were increased?
  3. How does the margin of error impact the width of the confidence interval?
  4. Why do we use a critical value from the z-distribution for large sample sizes?
  5. How can this confidence interval help in making business decisions?

Tip

Always double-check the critical value when constructing confidence intervals, as different confidence levels have different critical z-scores.

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Math Problem Analysis

Mathematical Concepts

Confidence Interval
Population Proportion
Sampling Distribution
Hypothesis Testing

Formulas

Point Estimate for Proportion: p-hat = x/n
Standard Error for Proportion: SE = sqrt((p-hat * (1 - p-hat)) / n)
Margin of Error: ME = z * SE
Confidence Interval: CI = p-hat ± ME

Theorems

Central Limit Theorem

Suitable Grade Level

Grades 10-12 or Introductory College Level