Math Problem Statement

A survey of 1010 college seniors working towards an undergraduate degree was conducted. Each student was asked, "Are you planning or not planning to pursue a graduate degree?" Of the 1010 surveyed, 658 stated that they were planning to pursue a graduate degree. Construct and interpret a 98% confidence interval for the proportion of college seniors who are planning to pursue a graduate degree. Question 28 options:

A)

(0.612, 0.690); we are 98% confident that the proportion of college seniors who are planning to pursue a graduate degree is between 0.612 and 0.690.

B)

(0.621, 0.680); we are 98% confident that the proportion of college seniors who are planning to pursue a graduate degree is between 0.621 and 0.680.

C)

(0.616, 0.686); we are 98% confident that the proportion of college seniors who are planning to pursue a graduate degree is between 0.616 and 0.686.

D)

(0.620, 0.682); we are 98% confident that the proportion of college seniors who are planning to pursue a graduate degree is between 0.620 and 0.682.

Solution

To solve this, we need to construct the 98% confidence interval for the population proportion pp. Here's the process:

Step 1: Calculate the sample proportion p^\hat{p}

p^=Number of successes (students planning graduate degrees)Total number of students surveyed=65810100.651\hat{p} = \frac{\text{Number of successes (students planning graduate degrees)}}{\text{Total number of students surveyed}} = \frac{658}{1010} \approx 0.651

Step 2: Determine the critical value zz^* for a 98% confidence level

For a 98% confidence level, the critical value zz^* corresponds to the 2.5% in each tail of the standard normal distribution. From the standard normal table: z2.33z^* \approx 2.33

Step 3: Calculate the standard error (SE)

SE=p^(1p^)n=0.651(10.651)10100.6510.34910100.22719910100.01499SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.651(1 - 0.651)}{1010}} \approx \sqrt{\frac{0.651 \cdot 0.349}{1010}} \approx \sqrt{\frac{0.227199}{1010}} \approx 0.01499

Step 4: Compute the margin of error (ME)

ME=zSE=2.330.014990.0349ME = z^* \cdot SE = 2.33 \cdot 0.01499 \approx 0.0349

Step 5: Construct the confidence interval

Confidence interval=p^±ME=0.651±0.0349\text{Confidence interval} = \hat{p} \pm ME = 0.651 \pm 0.0349 Lower bound=0.6510.03490.616\text{Lower bound} = 0.651 - 0.0349 \approx 0.616 Upper bound=0.651+0.03490.686\text{Upper bound} = 0.651 + 0.0349 \approx 0.686

Final Confidence Interval:

(0.616,0.686)(0.616, 0.686)

Interpretation:

We are 98% confident that the proportion of college seniors planning to pursue a graduate degree is between 0.616 and 0.686.

Correct Answer:

C) (0.616, 0.686); we are 98% confident that the proportion of college seniors who are planning to pursue a graduate degree is between 0.616 and 0.686.

Would you like me to provide further clarifications or calculations?


Here are 5 related questions for exploration:

  1. How does increasing the confidence level to 99% affect the width of the interval?
  2. What would happen to the confidence interval if the sample size was increased to 1500 students?
  3. How is the critical value zz^* determined for different confidence levels?
  4. Why is it necessary to assume the sample is random when constructing a confidence interval?
  5. How does the margin of error relate to the sample size and confidence level?

Tip: The larger your sample size, the smaller the margin of

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Proportions

Formulas

Sample Proportion: \( \hat{p} = \frac{x}{n} \)
Standard Error: \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \)
Margin of Error: \( ME = z^* \cdot SE \)
Confidence Interval: \( \hat{p} \pm ME \)

Theorems

Central Limit Theorem

Suitable Grade Level

College Statistics or Advanced High School