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In a survey of 500 college students, 250 of them said they had stayed up late more than 10 times in the past year. A 95% confidence interval is constructed based on the percentage of college students in this sample who stay up late more than 10 times. What is the lower limit for the interval? (round the answer to two decimal places)

Question

In a survey of 500 college students, 250 of them said they had stayed up late more than 10 times in the past year. A 95% confidence interval is constructed based on the percentage of college students in this sample who stay up late more than 10 times. What is the lower limit for the interval? (round the answer to two decimal places)

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Solution 1

To find the lower limit of the confidence interval, we first need to calculate the sample proportion (p̂) and the standard error (SE).

Step 1: Calculate the sample proportion (p̂) The sample proportion is calculated by dividing the number of students who stayed up late more than 10 times by the total number of students surveyed.

p̂ = 250 / 500 = 0.5

Step 2: Calculate the standard error (SE) The standard error for a proportion is calculated using the formula:

SE = sqrt [ p̂ * ( 1 - p̂ ) / n ]

where:

  • p̂ is the sample proportion
  • n is the sample size

SE = sqrt [ 0.5 * ( 1 - 0.5 ) / 500 ] = sqrt [ 0.5 * 0.5 / 500 ] = sqrt [ 0.0005 ] = 0.02236

Step 3: Calculate the margin of error (E) The margin of error for a 95% confidence interval is calculated using the formula:

E = Z * SE

where:

  • Z is the z-score, which is 1.96 for a 95% confidence interval
  • SE is the standard error

E = 1.96 * 0.02236 = 0.04386

Step 4: Calculate the lower limit of the confidence interval The lower limit of the confidence interval is calculated by subtracting the margin of error from the sample proportion.

Lower limit = p̂ - E = 0.5 - 0.04386 = 0.45614

So, the lower limit of the 95% confidence interval is approximately 0.46 (rounded to two decimal places).

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Solution 2

To construct a 95% confidence interval for a proportion, we use the formula:

CI = p ± Z * sqrt[(p(1-p))/n]

Where:

  • p is the sample proportion (in this case, 250/500 = 0.5)
  • Z is the Z-score for the desired confidence level (for a 95% confidence level, Z is approximately 1.96)
  • n is the sample size (in this case, 500)

Substituting these into the formula gives:

CI = 0.5 ± 1.96 * sqrt[(0.5(1-0.5))/500] = 0.5 ± 0.044

So the 95% confidence interval is (0.5 - 0.044, 0.5 + 0.044) = (0.456, 0.544)

Therefore, the lower limit for the interval is 0.456, or 45.6% when rounded to two decimal places.

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