The accompanying table lists the \'\'attribute\" ratings made by a random sampe
ID: 3054892 • Letter: T
Question
The accompanying table lists the ''attribute" ratings made by a random sampe of participants in a speed dating session. Each attribute ratings of five attributes. Use a .05 significance level to test the calaim that there is a difference between female attribute ratings and male attribute ratings. A. Identify the test statistic. B. Identify the P-value.
*first column female, second male
35 29
29 40
32 50
38 36
21 45
25 39
39 22
38 39
32 37
40 43
42 37
42 28
49 35
38 32
37 31
53 42
29 32
36 40
37 40
36 33
29 33
24 29
35 30
38 40
37 33
35 33
33 40
40 41
32 37
33 25
32 31
22 25
44 35
46 39
38 33
38 33
34 47
40 33
37 31
35 32
34 38
30 30
51 53
31 39
45 31
35 30
40 37
36 31
40 40
38 35
33 17
28 29
31 28
41 34
31 32
35 41
45 53
32 40
27 38
34 42
46 38
29 23
36 33
40 32
27 36
30 34
27 40
42 31
51 52
35 42
35 26
28 39
28 25
37 30
22 37
38 25
37 37
39 40
45 37
39 41
40 30
53 42
29 36
37 35
44 51
27 35
34 44
32 37
35 37
35 26
34 40
34 38
31 27
29 27
48 34
35 33
37 31
30 31
38 49
29 35
Explanation / Answer
here we use t-test with
null hypothesis H0:mean1=mean2 and
alternate hypothesis H1:mean1?mean2 ( two tailed test)
t=(mean1-mean2)/((sp*(1/n1 +1/n2)1/2)=0.264
and sp2=((n1-1)s12+(n2-1)s22)/n and with df is n=n1+n2-2
since two tailed p-value is more than alpha=0.05, so we fail to reject null hypothesis and conclude that there is no difference between female attreibutes ratings and male attributes ratings
sample mean s s2 n (n-1)s2 female 35.69 6.61 43.6921 100 4325.5179 male 35.44 6.79 46.1041 100 4564.3059 difference= 0.25 89.7962 200 8889.8238 sp2= 44.898 sp= 6.701 t= 0.264 one tailed p-value= 0.396 two tailed p-value= 0.792 one tailed critical t(0.05,n)= 1.653 two tailed critical t(0.05/2,n)= 1.972Related Questions
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