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The Population Variance and Standard Deviation - Speech or Presentation Example

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In the paper “The Population Variance and Standard Deviation” the author analyzes the population variance and standard deviation for the percentage rate of home ownership with 99% confidence. The percentage rates of home ownership were taken for 8 randomly selected states…
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The Population Variance and Standard Deviation
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The Population Variance and Standard Deviation
Section 7.1
14. A sociologist found that in a sample of 50 retired men, the average number of jobs they had during their lifetimes was 7.2. The population standard deviation is 2.1.
a) Find the best point estimate of the population mean.
b) Find the 95% confidence interval of the mean number of jobs.
c) Find the 99% confidence interval of the mean number of jobs.
d) Which is smaller? Explain why?
Solution
a) Best point estimate of the population mean is

b) At 95% confidence interval, the confidence coefficient  and the confidence interval is given by




c) At 99% confidence interval, the confidence coefficient 


d) The 95% confidence interval is smaller since the certainty of finding the true population mean is smaller compared to the 99% confidence interval.
20. The growing seasons for a random sample of 35 U.S. cities were recorded, yielding a sample mean of 190.7 days and a sample standard deviation of 54.2 days. Estimate the true mean population of the growing season with 95% confidence.
Solution
 and the confidence coefficient 


24. A pizza shop owner wishes to find the 95% confidence interval of the true mean cost of a large plain pizza. How large should the sample be if she wishes to be accurate to within $0.15? A previous study showed that the standard deviation of the price was $0.26.
Solution
At 95% confidence interval, the confidence coefficient 
Margin error is given by



25. If the variance of a national accounting examination is 900, how large a sample is needed to estimate the true mean score within 5 points with 99% confidence?





Section 7.2
14.  A recent survey of 8 social networking sites has a mean of 13.1 million visitors for a specific month. The standard deviation of the sample was 4.1 million. Find the 95% confidence interval of the true mean.
Solution



20. The number of unhealthy days based on the AQI (Air Quality Index) for a random sample of metropolitan areas is shown. Construct a 98% confidence interval based on the data
61 12 6 40 27 38 93 5 13 40
Solution



=29.23
Degrees of freedom = n-1 = 8-1 = 7, 


Confidence interval = 
C.I. (5.3, 67.2)



Section 7.3
12. It has been reported that 20.4% of incoming freshmen indicate that they will major in business or a related field. A random sample of 400 incoming college freshmen was asked their preference, and 95 replied they were considering business major. Estimate the true proportion of freshmen business majors with 98% confidence. Does your interval contain 20.4?
Solution



Confidence interval is given by

Hence C.I. (0.19,0.29) or C.I. (0.19×400, 0.29×400)= C.I.=(76,116)
Hence 20.4 is not contained in the interval



14. In a poll of 1000 likely voters, 560 say that the United States spends too little on fighting hunger at home. Find a 95% confidence interval for the true proportion of voters who feel this way.
Solution



Confidence interval

Hence C.I. (0.53,0.59)

16. A recent study indicated that 29% of the 100 women over age 55 in the study were widows. How large a sample must you take to be 90% confident that the estimate is within 0.05 of the true proportion of women over age 55 who are widows?
Solution






Section 7.3
8. Find the 90% confidence interval for the variance and standard deviation of seniors at Oak Park College if a sample of 24 students has a standard deviation of 2.4 years and the variable is normally distributed. What would be the 95% confidence interval?
Solution
This is chi-square distribution, for 90% confidence interval, we have  . for 0.05 and 0.95 and the d.f=23 is 34.172 and 13.091.




If the confidence interval was 95%, we have   for 0.025 and 0.975 and the d.f=23 is 11.61 and 38.08




12. The percentage rates of home ownership for 8 randomly selected states are listed below. Estimate the population variance and standard deviation for the percentage rate of home ownership with 99% confidence.
66.0 75.8 70.9 73.9 63.4 68.5 73.3 65.9
Solution



=11.07
At 99% confidence, and d.f=7,  we get as 0.99 and 20.28 Read More
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