marketing calcurate – Flashcards
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You have determined that a sample size of 250 will allow reasonable precision and confidence for your estimates of important population parameters. You will be conducting a telephone interview of university undergraduate students. After checking with university registration officials, you determine the campus directory to be 85% accurate. Further, you expect about 2% of the people you contact to be ineligible (e.g., graduate students), about 20% to decline your participation request, and about 30% not to be reachable even after trying at several different times on different days of the week. How many sampling elements should you include in the project?
answer
250/(1-BCI)(1-I)(1-R)(1-NC)
=250/(1-.15)(1-.02)(1-.2)(1-.3)
=536
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determining sample size(公式)
answer
n=(z^2/H^2)(est σ^2)
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Determine Sample Size:
You have been asked to determine the average amount that fishermen spend per year on food and lodging while on fishing trips in a certain state. Your estimate is to be within +/- $25 of the population mean, the confidence level is to be 95%, and the estimated standard deviation for the amount spent is $125 based on prior research.
answer
H=25
z=1.96
σ^2=125^2
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Determining Sample Size When Estimating Proportions(公式)
answer
n={z^2/H^2)π(1-π)
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Determining Sample Size When Estimating Proportions
You have been asked to determine the proportion of all out-of-state fishermen who took at least one overnight fishing trip in the past year. Your estimate is to be within +/- 2% of the population mean, the confidence level is to be 95%, and the best guess is that 25% of out-of-state respondents have taken at least one overnight fishing trip.
answer
H=2%
z=1.96
π=25%
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Response rates
You've conducted an email survey, sending out 500 surveys; the 500 sample elements have been classified as follows:
Usable questionnaires returned: 250
nSurveysSent: 500
Estimated eligibility percentage: .85
answer
250/(.85*500)=0.588
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What is the sampling error for means?
Researchers learned that the average number of visits to Avery Fitness over the last 30 days was 10 visits with a standard deviation of 7.3 from 198 responding members
We would like to establish a 95% confidence level (z = 1.96)
answer
1.96 (7.3/√198) = 1.0
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What is the confidence interval?
Researchers learned that the average number of visits to Avery Fitness over the last 30 days was 10 visits with a standard deviation of 7.3 from 198 responding members
We would like to establish a 95% confidence level (z = 1.96)
answer
10.0 - 1.0 ≤ x ≤ 10.0 + 1.0
