Question

For each of the following scenarios, give notation for the sample statistics and population parameter being estimated. Use the information provided to construct a 99% confidence interval. Write the interval in terms of population parameter. a. In a random sample of 100 families, 67 families have at least one dog.

- Sample statistic is _________________________

- Population parameter is _________________________

- Calculate the Standard Error -

Confidence Interval:

b. In the sample of 61 dog owners, the average veterinary expenditure per year is $100 with the standard deviation of $30. - Sample statistic is _________________________

- Population parameter is _________________________ -

Calculate the Standard Error -

Confidence Interval:

Answer 1

Ars n= 3632, Le 3125 Å= 2 3125 3632 0.860 0.860 â= 1-P = 0.140 And o n= 100, L=67 şample ostatistiski 뜸 67 100 = 0.67 P = 0.67 =) q= t-p = 0.33 Population parameter py True population proportion Standard Error: SEIP) pq 0.67x0,33 = 0.0470 100 SECO) = 0.0470 confidence interval CE 99.4 = 0.99, K=-C=0.01 22/2= Zogos 2,58 (from Z-table) ☆ SEC ) = 2.58 x000470 Margin of Error (EE 0.12126 ME=0:12 7412 S 99.% con fidend intervalf lower limit - ME = 0.67 -0.12 = 0.55 (11 = 0.55 Upper Limit = it ME 0167 +0.12 = 0,79 OL= 0,79 990% confidence interval : (0.55, 0.79) z

Ans a n= 61, I=100 S = 30 Som ple Sto tistics; I= $ 100 Population parametert M= Population mean Standard Error SE (W): 30 SE (X) = in NGI SE(X) = 3.0411 3.8411 tx 1-1 confidence interval: C = 99% =0.99, = 0.01 dt =n7=60 = +0.005,60 = 2-660 using R >qt 10.005,60, lowerotait=f) 12.660203 Margin of Error (E) = til,df * 2.660 * 7.8411 = Margin of Error (ME)= .10.22 SECH - 10.2017326 99% confidence interval for e 으 lower - ME= 100 10.22 = 89.78 limit Lk = 89.78 110.22 Upper Limit = TE+ME = 100 + 10,225 UL=110.22 99% confidence interval ; (09.78, 110:22)