A researcher wishes to estimate with 95% confidence the proportion of adults who have high speed internet access. Her estimate must be accurate within 4% of the true proportion.

Question

A researcher wishes to estimate with 95% confidence the proportion of adults who have high speed internet access. Her estimate must be accurate within 4% of the true proportion.

1. Find the minimum sample size needed using a prior study that found that 48% of the respondents said they have high speed internet access
2. No preliminary estimate is available. Find the minimum sample size needed.

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4 answers
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Answer 1

MathGuru answered
12 years ago

Formula to find sample size:

n = [(z-value)^2 * p * q]/E^2
... where n = sample size, z-value is found using a z-table for 95% confidence, p = .48, q = 1 - p, ^2 means squared, * means to multiply, and E = .04 (or 4%).

For 2): use p = .5 (when no value is stated in the problem), q = 1 - p

I hope this will help get you started.

Answer 2

Ari answered
11 years ago

1) A group of scientists created 150 trials to measure whether electric shock treatment could cure paranoid delusions. Of these trials, 52 were successful. Find the margin of error E that corresponds to a 95% confidence level. The critical value for 95% confidence level is 1.96.

2) Find the minimum sample size that should be chosen to assure that the proportion estimate p will be within the required margin of error, .06. Use a 95% confidence interval and a population proportion of .7. The critical value for a 95% confidence level is 1.96

3) Find the test statistic for the following proportion: A collection of 500 randomly selected teachers revealed that 61% felt that all students should be required to take algebra in high school.

4) Employees in a large computer firm claim that the mean salary of the firm™s programmers is less than that of its competitors. The competitor™s salary is $47,000. A random sample of 30 of the firm™s programmers has a mean salary of $46,500 with a standard deviation of 5500. Calculate the test statistic for the hypothesis: Ho: mean >= 47000, H1: mean < 47000

Answer 3

Ari answered
11 years ago

Find the minimum sample size that should be chosen to assure that the proportion estimate p will be within the required margin of error, .06. Use a 95% confidence interval and a population proportion of .7. The critical value for a 95% confidence level is 1.96

Find the test statistic for the following proportion: A collection of 500 randomly selected teachers revealed that 61% felt that all students should be required to take algebra in high school.

Employees in a large computer firm claim that the mean salary of the firm™s programmers is less than that of its competitors. The competitor™s salary is $47,000. A random sample of 30 of the firm™s programmers has a mean salary of $46,500 with a standard deviation of 5500. Calculate the test statistic for the hypothesis: Ho: mean >= 47000, H1: mean < 47000

Answer 4

Explain Bot answered
1 year ago

1. To find the minimum sample size needed with a prior estimate available, we can use the formula for sample size calculation for proportions:

n = (Z^2 * p * (1-p)) / E^2

Where:
- n is the required sample size
- Z is the z-score corresponding to the desired level of confidence (in this case, 95%), which is approximately 1.96
- p is the prior estimate (48% or 0.48 in decimal form)
- E is the desired margin of error (4% or 0.04 in decimal form)

Substituting the given values into the formula, we have:

n = (1.96^2 * 0.48 * (1-0.48)) / 0.04^2
n = (3.8416 * 0.48 * 0.52) / 0.0016
n = 0.9867 / 0.0016
n ≈ 616.69

So, the researcher would need a minimum sample size of approximately 617 participants to estimate the proportion of adults with high-speed internet access with a 95% confidence interval and a margin of error of 4% based on the prior estimate of 48%.

2. Without a preliminary estimate available, we can assume the most conservative estimate, which is p = 0.5 (50% in decimal form). This is because when no information about the proportion is available, assuming p = 0.5 provides the largest sample size requirement, giving the most conservative estimate.

Using the same formula as above:

n = (Z^2 * p * (1-p)) / E^2

Substituting the values:

n = (1.96^2 * 0.5 * (1-0.5)) / 0.04^2
n = (3.8416 * 0.5 * 0.5) / 0.0016
n = 0.9604 / 0.0016
n ≈ 600.25

Therefore, a minimum sample size of approximately 601 participants would be needed to estimate the proportion of adults with high-speed internet access with a 95% confidence interval and a margin of error of 4% without a preliminary estimate available.