The dataset lists the auction success rate for Item A (Column: SuccessRate-A), a

Question

The dataset lists the auction success rate for Item A (Column: SuccessRate-A), and for Item B (Column: SuccessRate-B) at eBay. You wonder if both rates are equal to 0.70, respectively. Please write a null hypothesis and alternative hypothesis for each item. Then, use SPSS for one sample t-test. Please briefly interpret the testing result.

Day SuccessRate-A SuccessRate-B 1 0.7895 0.5789 2 0.8800 0.8800 3 0.8182 0.7273 4 0.6818 0.6364 5 0.6957 0.6957 6 0.7959 0.6531 7 0.6462 0.5077 8 0.6842 0.6184 9 0.6744 0.6124 10 0.6522 0.5725 11 0.7059 0.6975 12 0.6595 0.6293 13 0.7371 0.6743 14 0.7213 0.6995 15 0.6828 0.6564 16 0.5613 0.5032 17 0.7394 0.7183 18 0.7117 0.6396 19 0.7300 0.6900 20 0.5915 0.5362 21 0.6907 0.6667 22 0.6770 0.6433 23 0.7032 0.6643 24 0.7554 0.7038 25 0.7574 0.7346 26 0.6798 0.6601 27 0.5342 0.4958 28 0.7872 0.7748 29 0.7952 0.7714 30 0.7632 0.7394 31 0.7708 0.7569 32 0.7763 0.7451 33 0.7519 0.7245 34 0.7145 0.6598 35 0.7167 0.6771 36 0.7657 0.7336 37 0.7648 0.7378 38 0.7425 0.7180 39 0.7373 0.7066 40 0.6881 0.6823

Explanation / Answer

Descriptive statistics:

SuccessRate-A

count 40

mean 0.718263

sample variance 0.004500

sample standard deviation 0.067081

minimum 0.5342

maximum 0.88

range 0.3458

null, Ho: =0.7

alternate, H1: !=0.7

level of significance, = 0.05

from standard normal table, two tailed t /2 =2.023

since our test is two-tailed

reject Ho, if to < -2.023 OR if to > 2.023

we use test statistic (t) = x-u/(s.d/sqrt(n))

to =0.718-0.7/(0.067081/sqrt(40))

to =1.697

| to | =1.697

critical value

the value of |t | with n-1 = 39 d.f is 2.023

we got |to| =1.697 & | t | =2.023

make decision

hence value of |to | < | t | and here we do not reject Ho

p-value :two tailed ( double the one tail ) - Ha : ( p != 1.6971 ) = 0.0976

hence value of p0.05 < 0.0976,here we do not reject Ho

ANSWERS

---------------

null, Ho: =0.7

alternate, H1: !=0.7

test statistic: 1.697

critical value: -2.023 , 2.023

decision: do not reject Ho

p-value: 0.0976

----------------------------------------------------------------------------------------

Descriptive statistics: SuccessRate-B

count 40

mean 0.673065

sample variance 0.006242

sample standard deviation 0.079005

minimum 0.4958

maximum 0.88

range 0.3842

null, Ho: =0.7

alternate, H1: !=0.7

level of significance, = 0.05

from standard normal table, two tailed t /2 =2.023

since our test is two-tailed

reject Ho, if to < -2.023 OR if to > 2.023

we use test statistic (t) = x-u/(s.d/sqrt(n))

to =0.67306-0.7/(0.079005/sqrt(40))

to =-2.157

| to | =2.157

critical value

the value of |t | with n-1 = 39 d.f is 2.023

we got |to| =2.157 & | t | =2.023

make decision

hence value of | to | > | t | and here we reject Ho

p-value :two tailed ( double the one tail ) - Ha : ( p != -2.1566 ) = 0.0373

hence value of p0.05 > 0.0373,here we reject Ho

ANSWERS

---------------

null, Ho: =0.7

alternate, H1: !=0.7

test statistic: -2.157

critical value: -2.023 , 2.023

decision: reject Ho

p-value: 0.0373

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