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Suppose Facebook has launched a new fact-checking feature. It conducts a survey among its user base, wherein it asks them the question ‘Do you find the fact-checking feature useful?’. Out of the 2,500 respondents, 53.2% believe that the feature is useful. If you define X as the proportion of people that find the feature useful, the frequency distribution for X would be as shown in the table given below. X Frequency1 1,3300 1,170Question 2/3MandatorySampling and EstimationSuppose the sample standard deviation is known to be 0.6, what is the interval in which the mean of the sampling distribution of sample size 2,500 will belong at a 99% confidence level?(0.456, 0.608)(0.572, 0.588)(0.493, 0.571)(0.501, 0.563)

Question

Suppose Facebook has launched a new fact-checking feature. It conducts a survey among its user base, wherein it asks them the question ‘Do you find the fact-checking feature useful?’. Out of the 2,500 respondents, 53.2% believe that the feature is useful. If you define X as the proportion of people that find the feature useful, the frequency distribution for X would be as shown in the table given below. X Frequency1 1,3300 1,170Question 2/3MandatorySampling and EstimationSuppose the sample standard deviation is known to be 0.6, what is the interval in which the mean of the sampling distribution of sample size 2,500 will belong at a 99% confidence level?(0.456, 0.608)(0.572, 0.588)(0.493, 0.571)(0.501, 0.563)

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Solution

To find the interval in which the mean of the sampling distribution of sample size 2,500 will belong at a 99% confidence level, we need to use the formula for the confidence interval:

CI = X ± Z * (σ/√n)

where:

  • X is the sample mean
  • Z is the Z-score (which is 2.576 for a 99% confidence level)
  • σ is the standard deviation
  • n is the sample size

First, we need to calculate the sample mean. Given that 53.2% of the 2,500 respondents find the feature useful, the sample mean is 0.532.

Next, we plug the values into the formula:

CI = 0.532 ± 2.576 * (0.6/√2500)

This simplifies to:

CI = 0.532 ± 0.031

So, the interval in which the mean of the sampling distribution of sample size 2,500 will belong at a 99% confidence level is (0.501, 0.563). Therefore, the correct answer is (0.501, 0.563).

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