8) The average annual published cost of tuition, room and board at private, 4 ye

Question

8) The average annual published cost of tuition, room and board at private, 4 year university for the academic year 2017-2018 is $46,950 according to the "Trends in College Pricing 2017'9 report of the College Board. A group of financial aid administrators believe that the average cost is lower. A study conducted using a sampling of 25 private universities showed that the average cost per year is $42,874 with a standard deviation of $1,61 3 8 a) How are the means distributed? Explain. b) Find the 95% confidence interval for the average college cost according to the 6 administrators c) How would this be distributed if the sample size ere increased to 225? d) Find the confidence interval with a sample size of 225. e) How many would you need to sample is you have an allowable error of $690? 8 9) According to the Census Bureau, the homeownership vacancy rate in the US is 1.8%" in the 4th quarter of 2016. We sampled 2,00 homeowners and found 8 vacant homes. Based on the information above a) Find a 90% confidence interval for homeownership vacancies b) Based on the data, is this a representative sample if the census data is correct? 80S

Explanation / Answer

Question 8

Here means of sample size 25 will be distributed approximately normally and can be represented by student's t distribution.

(b) 95% confidence interval = x +- tdf,0.05 se0

Here x = $ 42874

s = 1618

so stadard error of the sample mean se0 = s/ sqrt(n) = 1618/ sqrt(25) = 323.6

dF = 24 and alpha = 0.05

so t0.05, 24 = 2.0639

so 95% confidence interval = x  +- tdf,0.05 se0 = 42874 +- 2.0639 * 323.6 = (42206.12, 43541.88)

(c) It would be distriubted normally when sample size is 225 as it is greater than 30.

(d) 95% confidence interval = x  +- tdf,0.05 se0   

here stadard error of the sample mean se0 = s/ sqrt(n) = 1618/ sqrt(225) = 107.87

95% confidence interval = x  +- tdf,0.05 se0 = 42874 + 2.0639 * 107.87 = (42651.37, 43096.63)

(e) Here allowable error = $ 690 = Critical Z statistic * Standard errr of the sample mean

690 = 1.96 * 1618/ sqrt(n)

sqrt(n) = 1.96 * 1618/ 690 = 4.596

n = 21.12

so n would be 22

Question 9

(a) Here sample proportion p^ = 8/200 = 0.04

90% confidence interval = p^ + - Z90% sqrt[p^ * q^ /N]

= 0.04 +- 1.96 * sqrt [0.04 * 0.96/200]

= 0.04 +- 0.0272

= (0.0128, 0.0672)

(b) yes, the confidence interval encomasses tha value of 1.8% or 0.018 in the confidence interval so we shall say the sample represents the populaton.

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