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Chirag, Master's Degree
Category: Calculus and Above
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Experience:  I have a Master's degree in Engineering and a very wide teaching experience of more than 25 years at various levels.
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# You answered similar questions previously but the data has

### Resolved Question:

You answered similar questions previously but the data has slighly changed. Can you please provide details as to how you worked the problem? Thx

#1 Find the sample size needed to estimate the percentage of adults who have consulted fortune tellers. Use a 0.03 margin of error, use a confidence of 98%, and use the results from a prior Pew Research Center Poll suggesting that 15% of adults have consulted fortune tellers.

#2SAMPLE SIZE USING SAMPLE DATA- Refer to DATA 16 in Appendix B and find the maximum and minimum earthquake magnitudes, then use those values with the range rule of thumb to estimate σ.
How many earthquakes must you randomly select and test if you want to be 95% confident that the sample mean magnitude is 0.2 of the true population mean µ? The standard deviation of the 50 earthquake magnitudes in Data Set 16 is 0.587. Find the required sample size by using the standard deviation of the sample as an estimate of σ. Which of the two results is likely to be better?

#3 Use the ages (years) of the presidents at the times of their inaugurations. Treating the data as sample, construct a 98% confidence interval estimate of the standard deviation of the population of all such ages. 57,61,57,57,58,57,61,54,68,49,64,48,65,52,46,54,49,47,55,54,42,51,56,55,51,54,51,60,62,43,55,56,52,69,64,46,54,47

#4 You want to estimate σ for the population of waiting times at Mcdonalds drive up windows, and you want to be 95% confident that the sample standard deviation is within 20% of σ . Find the minimum sample size. Isthis sample size practical? (sameple size for 20% is 48)
Submitted: 1 month ago.
Category: Calculus and Above
Expert:  Chirag replied 1 month ago.
Hi, Welcome and Thanks for using Just Answer

I am working on this and will post the answers as soon as ready.

Thanks.
Expert:  Chirag replied 1 month ago.
Here they are ...

(1)

Confidence Level % = 98
z- score = 2.3263
Population Proportion, p = 0.15
q = 1 - p = 0.85
Error, E = 0.03
N = (z/E)^2 * pq = (2.3263/0.03)^2 * 0.15 * 0.85 = 766.65
Sample size = 767

(2)

(a) Using range rule of thumb:
Range = Maximum value – Minimum value = 2.65 – 0 = 2.65
σ = Range/4 = 2.65/4 = 0.6625
Confidence Level % = 95
z- score = 1.96
Population SD, σ = 0.6625
Error, E = 0.20
N = (z * σ / E)^2 = (1.96 * 0.6625/0.2)^2 = 42.15
Sample size = 43
(b) Using the sample σ = 0.587
Confidence Level % = 95
z- score = 1.96
Population SD, σ = 0.587
Error, E = 0.20
N = (z * σ / E)^2 = (1.96 * 0.587/0.2)^2 = 33.09
Sample size = 34
Larger sample size estimate of 43 is preferred.

(3)

n = 38
x-bar = 54.76315789
s = 6.561287821
% = 98
Standard Error, SE = s/n = 6.56128782071684/√38 = 1.064381435
Degrees of freedom = n - 1 = 38 -1 = 37
t- score = 2.4314474
Width of the confidence interval = t * SE = 2.43144740046467 * 1.06438143472465 = 2.587987473
Lower Limit of the confidence interval = x-bar - width = 54.7631578947368 - 2.58798747256411 = 52.1752
Upper Limit of the confidence interval = x-bar + width = 54.7631578947368 + 2.58798747256411 = 57.3511
The 98% confidence interval is [52.18, 57.35]

(4)

The sample size of 48 appears to be practical since it is greater than 30 and provides a reliable basis for other computations.

I hope that helps. Let me know if you need any clarifications. Please Rate My Answer (Tip If Happy).

Thanks.
Chirag, Master's Degree
Satisfied Customers: 11960
Experience: I have a Master's degree in Engineering and a very wide teaching experience of more than 25 years at various levels.
Customer: replied 1 month ago.

Thank you! excellent and quick work!

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