In Study Design 2, Super Sneaker Company drew at random two groups of 16 high sc

Question

In Study Design 2, Super Sneaker Company drew at random two groups of 16 high school students from the Halifax school district database. After obtaining their shoe sizes, the company manufactured 16 pairs of shoes for group 1, each pair with both soles constructed from material A, and 16 pairs of shoes for group 2, each pair with both soles constructed from material B. After 3 months, the amount of sole wear in each shoe was recorded in standardized units, as in the first design. The researcher in charge of this design wanted cnily one measurement of shoe wear per student. To get only one measurement per student, several options are possible: (T) measuring only left shoe wear (or only right shoe wear). (II) measuring left or right shoe wear at random. (III) averaging the left and right shoe wear for each pair, The researcher chose option III. Why? Incorrect Right shoe wear is always less than left shoe wear Incorrect Right shoe wear is always greater than left shoe wear Incorrect Average of wear on left and right foot will be more variable than wear on either foot individually. Correct: Option III uses information from both the left and right foot equally. 1 pt(s)) are correct. Using option I1I, the following data were used to test the hypothesis that wear for material B was greater than wear for material A 17.66 14.11 12.43 15.76 13.18 12.90 14.56 14.42 16.71 12.65 13.45 17.26 14.84 14.13 16.60 16.22 16.27 16.65 15.42 14.86 16.95 15.73 14.54 16.14 16.11 15.22 14-23 13.47 18.32 15.10 15.36 16.87 Group 1 -MaterialA Group 2 Material B A null hypothesis for the test is: Correct: UB- HA 1 ot(s)) Your receipt no. i 158-4174eviosus Trles An alternative hypothesis is: Correct: B- incorrect .->0 Incorrect X A-XBCO Your recelipt no. is80Prexious Tries

Explanation / Answer

t-statistic:

Let  d = differences between the pairs of data, then ¯dd¯ = mean of these differences.

The test statistics is: t=¯d0sd/nt=d¯0sd/n

degrees of freedom = n - 1
where n denotes the number of pairs or the number of differences.

Test statistics t=1.6972

Degrees of freedom=n1+n2-2

=30

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