Below is a set of data for home sale prices (in thousands of dollars) collected

Question

Below is a set of data for home sale prices (in thousands of dollars) collected in a small community. Please examine the data and find the mean, median and standard deviation of the data set 50, 75, 100, 125, 150, 150, 200, 250, 250, 300 Median = _____ Mean = _____ Standard deviation = _____ (please Show your calculation below.) Suppose that a researcher took a simple random sample of 300 households last year to estimate the monthly expense on department store shopping and found that the mean expense for the sample is $182 and the standard deviation is $55. Please use the given information to calculate the 95% confidence interval for households' average monthly spending on department store shopping

Explanation / Answer

(1) there are 10 observation here,

to find the median we arrange the observation in ascending or descending order and middle observation is median.

if number of observation n is odd then (n/2 + 1)th observation will be median. if n is even then average of n/2 and (n/2 + 1)th observation will be median.

here n=10 so average 5th and 6th observation will be median

medain=(150+150)/2=150

sum=50+75+...+250+300=1650

n=10 , mean=sum/n=1650/10=165

standard deviation=sqrt(sum(x-mean)2/n)=sqrt(61500/10)=78.42

(2) 1-alpha)*100% confidence interval for mean=mean± z(alpha/2)*sd/sqrt(n)

95% confidence interval for mean=mean±z(0.05/2)*sd/sqrt(n)=182±1.96*55/sqrt(300)=182±6.22=(175.78,188.22)

s.n. x x-mean (x-mean)2 1 50 -115 13225 2 75 -90 8100 3 100 -65 4225 4 125 -40 1600 5 150 -15 225 6 150 -15 225 7 200 35 1225 8 250 85 7225 9 250 85 7225 10 300 135 18225 sum= 1650 0 61500 mean= sum/n= 1650/10= 165 n=10

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