A 2007 Carnegie Mellon University study reported that onlineshoppers were willing to pay, on average, more than an extra $0.60on a $15 purchase in order to have better online privacyprotection.
A sample of ?=22n=22 online shoppers was taken, and each was askedhow much extra would you pay, on a $15 purchase, for better onlineprivacy 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? Usealpha = 0.05.   ? yes no  

(b) Determine the ?P-value for this Normality test, to threedecimal places.
?=P=

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


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

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

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(f) Based on the above calculations, we should  ? rejectnot reject  the null hypothesis. Use alpha = 0.05