Can someone please help me answer these questions? I have tried every method I know and the computer still keeps marking the answers as incorrect. Thanks for the help!
1) If the number of 18-34 year olds sampled is equal to the value identified in part "a"(value=157), what is the standard deviation of the percentage that uses an online tax program (to 2 decimals)? [the percentage given of all 18-34 year olds who use online tax program was 16%]
2) If the number of 35-44 year olds sampled is equal to the value identified in part "b" (value=209, what is the standard deviation of the percentage that uses an online tax program(to 2 decimals)? [the percentage given of all 35-44 year olds who use online tax program was 12%)
Standard deviation = √npq
1) n = 157
p = .16 (for 16%)
q = 1 - p = 1 - .16 = .84
2) n = 209
p = .12 (for 12%)
q = 1 - p = 1 - .12 = .88
Substitute and calculate.
c. How many 65+ year olds must be sampled to find an expected number of at least 25 that use an online tax program to prepare their federal income tax return?.
25/.02 =
1250
To answer these questions and calculate the standard deviation of the percentage of individuals who use an online tax program, we need some additional information. Specifically, we need the sample sizes (number of individuals sampled) for the 18-34 year olds and the 35-44 year olds.
Once we have the sample sizes, we can calculate the standard deviation using the following formula:
Standard deviation = √(p * (1-p) / n)
Where:
- p is the percentage of individuals who use an online tax program, given in the question.
- n is the sample size.
Let's assume that we have the sample sizes for both age groups. Now we can proceed to calculate the standard deviation for each question.
1) For the 18-34 year olds:
- Sample size (n) = 157 (given in part "a").
- Percentage (p) = 16% = 0.16 (given in the question).
Standard deviation = √(0.16 * (1-0.16) / 157)
= √(0.16 * 0.84 / 157)
= √(0.1344 / 157)
≈ √0.000856 GT
Therefore, the standard deviation of the percentage for the 18-34 year olds is approximately 0.029.
2) For the 35-44 year olds:
- Sample size (n) = 209 (given in part "b").
- Percentage (p) = 12% = 0.12 (given in the question).
Standard deviation = √(0.12 * (1-0.12) / 209)
= √(0.12 * 0.88 / 209)
= √(0.1056 / 209)
≈ √0.000505 GT
Therefore, the standard deviation of the percentage for the 35-44 year olds is approximately 0.022.
Please note that these calculations assume simple random sampling without replacement and that the sample is representative of the population.
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