Question

6. Confidence intervals A survey of 90 randomly selected families showed that 40 owned at least one television set, Find a 95% confidence interval for the proportion of all families that own at least one television set. a. b. A random sample of 20 automobiles has a pollution by-product release standard deviation of 2.3 ounces when 1 gallon of gasoline is used. Find a 90% confidence interval for the population standard deviation of the pollution by-product release. An irate patient complained that the cost of a doctor's visit was too high, so she randomly sampled 20 other patients and found that the mean amount of money they spent on visits after insurance was $44.80. The standard deviation of that sample is $3.53. Find a 95% confidence interval for the population mean amount of money spent at this doctor's office. c.

Answer 1

Ans 6

a.

Normal approximation to the binomial calculation:

Standard error of the mean = SEM = ?x(N-x)/N³ = 0.052

? = (1-CL)/2 = 0.025

Standard normal deviate for ? = Z_? = 1.960

Proportion of positive results = P = x/N = 0.444

Lower bound = P - (Z_?*SEM) = 0.342

Upper bound = P + (Z_?*SEM) = 0.547

b)

Find high end confidence interval value ahigh end 1- a/2 high end 1-0.1/2 ahigh end = 0.95 Find high end y value for 0.95 X2095 = 10.117 <一Value can be found on Excel using-CHINV(09519) High End = Square Root(n-1)s2/X21-12) High End (19)%5.29/10.117) High End v100.51/10.117 High End 9.934763269744 High End 3.1519 1.826 σ 3.1519 <--This is our 90% confidence interval

c)

CI [43.253, 46.347]

Calculation M 44.8 t 1.96 SMV(3.532/20) 0.79 μ-44.8 μ-44.8 1.96*0.79 1.547