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 ?=22n=22 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 ?P-value for this Normality test, to three decimal places.
?=P=

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(c) Choose the correct statistical hypotheses.
A. ?0:?>0.60??:?=0.60H0:μ>0.60HA:μ=0.60
B. ?0:?⎯⎯⎯⎯⎯=0.60,??:?⎯⎯⎯⎯⎯<0.60H0:X¯=0.60,HA:X¯<0.60
C. ?0:?>0.60,??:?<0.60H0:μ>0.60,HA:μ<0.60
D. ?0:?⎯⎯⎯⎯⎯=0.60,??:?⎯⎯⎯⎯⎯>0.60H0:X¯=0.60,HA:X¯>0.60
E. ?0:?=0.60,??:?≠0.60H0:μ=0.60,HA:μ≠0.60
F. ?0:?=0.60??:?>0.60H0:μ=0.60HA:μ>0.60


(d) Determine the value of the test statistic for this test, using two decimals in your answer.
Test Statistic =

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(e) Determine the ?P-value for this test, enter your answer to three decimals.
?=P=

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

Answer 1

Using Minitab, (Stat -> Basic Statistics -> Normality test -> Kolmogorov-Smirnov test), we get the following output -

P-value < 0.010,

a) Since P-value < 0.05, so data does not follow an approximately Normal distribution.

c) - f)