3. a. Distinguish between a point estimator and interval estimate. b. A recent s

Question

3. a. Distinguish between a point estimator and interval estimate.  

b. A recent survey was conducted to find out the amount of money households spent on Christmas gifts in 2014. A random sample of 3600 households were asked to provide estimates of their expenditure on Christmas gifts for 2014. A summary of the expenditure for the random sample is shown below.

i. Calculate the sample mean and sample standard deviation.  

ii. Identify the estimator used to estimate the population mean

iii. Name and define desirable property of this estimator

iv. Compute the 95% confidence interval for the household mean expenditure on Christmas gifts in 2014.     

v. Merchants consider a good Christmas sale when household spends on average more than EC$2000 dollars. Test at the 5% level of significance whether merchants had a good Christmas sale, on Christmas gifts.    

3600 X Xi 5,040,000 i-1

Explanation / Answer

a)

point estimate estimate popn parameter. eg avg, proportion etc.

interval estimate gives a range of values in which popn parameter lies.

b) i) mean = sum(Xi)/n =5040000/3600 1400 var =sum(Xi-xbar)^2/(n-1) =439569/(3600-1) 122.1364268 sd =sqrt(var) = 11.05153504 ii) average iii) sum of all observation from mean is zero. iv) 95% ci = mean+-z(a/2)*(Sd/sqrt(n)) lower =1400-1.96*(11.05/SQRT(3600)) 1399.639 upper =1400+1.96*(11.05/SQRT(3600)) 1400.361 v) ho u = 2000 h1: u > 2000 z = (mean-u)/(sd/sqrt(n)) z = =(1400-2000)/(11.05/SQRT(3600)) -3257.918552 since, z > z(a/2), I reject ho and conclude that u > 2000

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