Question

A 2007 Carnegie Mellon University study reported that online shoppers were willing to pay, on average, more than an extra $0.60 on a $15 purchase in order to have better online privacy protection.
A sample of n=22n online shoppers was taken, and each was asked how much extra would you pay, on a $15 purchase, for better online privacy protection?'' The data is given below, in $'s.

0.79,0.41,0.67,0.67,0.83,0.76,0.55,0.92,0.61,0.57,0.54,1.25,0.70,0.85,0.59,0.59,0.90,0.67,0.62,0.67,0.44,0.500.79,0.41,0.67,0.67,0.83,0.76,0.55,0.92,0.61,0.57,0.54,1.25,0.70,0.85,0.59,0.59,0.90,0.67,0.62,0.67,0.44,0.50

(a) Do the data follow an approximately Normal distribution? Use alpha = 0.05.  ? yes no

(b) Determine the PP-value for this Normality test, to three decimal places.
P=

(c) Choose the correct statistical hypotheses.
A. H0:μ=0.60HA:μ>0.60H0:μ=0.60HA:μ>0.60
B. H0:X¯¯¯¯=0.60,HA:X¯¯¯¯<0.60H0:X¯=0.60,HA:X¯<0.60
C. H0:μ>0.60HA:μ=0.60H0:μ>0.60HA:μ=0.60
D. H0:X¯¯¯¯=0.60,HA:X¯¯¯¯>0.60H0:X¯=0.60,HA:X¯>0.60
E. H0:μ=0.60,HA:μ≠0.60H0:μ=0.60,HA:μ≠0.60
F. H0:μ>0.60,HA:μ<0.60H0:μ>0.60,HA:μ<0.60


(d) Determine the value of the test statistic for this test, using two decimals in your answer.
Test Statistic =
(e) Determine the P-value for this test, enter your answer to three decimals.
P=

(f) Based on the above calculations, we should  ? reject not reject  the null hypothesis. Use alpha = 0.05

Answer 1

a = A Normality test by r software

And codes are given below,

> X

[1] 0.79 0.41 0.67 0.67 0.83 0.76 0.55 0.92 0.61 0.57 0.54 1.25 0.70 0.85 0.59 0.59 0.90 0.67 0.62

[20] 0.67 0.44 0.50

> shapiro.test(X)

        Shapiro-Wilk normality test

data:  X

W = 0.91685, p-value = 0.065 

b=  p-value = 0.065 

From the output, the p-value > 0.05 implying that the distribution of the data are not significantly different from normal distribution. In other words, we can assume the normality.

C =

Hypothesis =

H₀ :μ=0.60 V/S H₁ :μ>0.60