In: Math
Count | iPhone User Id | Price Willing to Pay |
1 | 101 | 1150 |
2 | 204 | 800 |
3 | 205 | 1050 |
4 | 405 | 1400 |
5 | 701 | 1050 |
6 | 105 | 1350 |
7 | 98 | 700 |
8 | 12 | 1450 |
9 | 37 | 800 |
10 | 55 | 650 |
11 | 68 | 750 |
12 | 31 | 1200 |
13 | 90 | 500 |
14 | 92 | 950 |
15 | 447 | 1050 |
16 | 778 | 1150 |
You are hired by Google to research how much people are willing to pay for a new cell phone in US. They are especially interested to know if their new phone, Pixel 3, should be priced similarly to Apple’s iPhone Xs. Google believes that there is a difference between what Android and iPhone users are willing to pay for high-end phones. You are hired to answer this question. Part I: To analyze iPhone users your team randomly selects 16 individuals. See attached data file.
a) Compute sample mean and sample standard deviation for iPhone users
b) Compute 5-number summary for iPhone users
c) Find the 90% confidence interval for the average phone price iPhone users are willing to pay. How do you interpret it?
sample mean=sum/n==3429/16=214.3125
sample standard deviation=sqrt(sum((x-)2)/(n-1))=sqrt(866799.4375/(16-1))=240.39
Count | iPhone User Id(x) | (x-214.3125)2 | x-214.3125 |
1 | 101 | -113.31 | 12839.72 |
2 | 204 | -10.31 | 106.35 |
3 | 205 | -9.31 | 86.72 |
4 | 405 | 190.69 | 36361.72 |
5 | 701 | 486.69 | 236864.72 |
6 | 105 | -109.31 | 11949.22 |
7 | 98 | -116.31 | 13528.60 |
8 | 12 | -202.31 | 40930.35 |
9 | 37 | -177.31 | 31439.72 |
10 | 55 | -159.31 | 25380.47 |
11 | 68 | -146.31 | 21407.35 |
12 | 31 | -183.31 | 33603.47 |
13 | 90 | -124.31 | 15453.60 |
14 | 92 | -122.31 | 14960.35 |
15 | 447 | 232.69 | 54143.47 |
16 | 778 | 563.69 | 317743.60 |
sum= | 3429 | 0 | 866799.4375 |
n= | 16 | ||
sample mean=sum/n= | 214.3125 | ||
sample sd= | 240.3884963 |
(b) five number summery using R is given as
x=c(101,204,205,405,701,105,98,12,37,55,68,31,90,92,447,778)
> summary(x)
Min. 1st Qu. Median 3rd Qu. Max.
12.00 64.75 99.50 255.00 778.00
(c) (1-alpha)*100% confidence interval for population mean=sample mean±t(alpha/2,n-1)*s/sqrt(n)
90% confidence interval =214.3125±t(0.1/2, n-1)*240.39/sqrt(16)=214.3125±1.75*240.39/sqrt(16)=
214.3125±105.35=(108.96,319.67)
n= | 16 |
sample mean= | 214.3125 |
s= | 240.39 |
t-value | margin of error | lower limit | upper limit | |
95% confidence interval | 2.13 | 128.09 | 86.22 | 342.41 |
99% confidence interval | 2.95 | 177.09 | 37.22 | 391.40 |
90% confidence interval | 1.75 | 105.35 | 108.96 | 319.67 |