Answer:
Since the calculated value of t = 0.01145 does not fall in the critical region ║t║≥ t (0.025,19) = 2.093 so we accept H0 the data suggests that after training people can correctly predict coin toss outcomes better than the 50 percent expected by chance guessing alone.
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Step-by-step explanation:
Researchers want to see whether training increases the capability of people to correctly predict the outcomes of coin tosses. Each of twenty people is asked to predict the outcome (heads or tails) of 100 independent tosses of a fair coin. After training, they are retested with a new set of 100 tosses. (All 40 sets of 100 tosses are independently generated). The numbers correct for each of the 20 people were as follows:
a. Do the data suggest that after training people can correctly predict coin toss outcomes better than the 50 percent expected by chance guessing alone? Give appropriate statistical evidence to support your claim.
Score Before Training Score After Training
(number correct) (number correct) Difference d²
46 61 -15 225
48 62 - 14 196
50 53 -3 9
54 46 8 64
54 50 4 16
54 52 2 4
54 53 1 1
54 59 -5 25
54 60 -6 36
54 61 -7 49
55 55 0 0
56 59 -3 9
57 56 1 1
58 50 8 64
58 56 2 4
61 58 3 9
61 64 -3 9
63 57 6 36
64 61 3 9
<u> 65 54 11 121 </u>
<u>∑ -7 887 </u>
The degrees of freedom = n-1= 20-1= 19
The significance level is 0.05
The test statistic is
t= d`/sd/√n
The critical region is ║t║≥ t (0.025,19) = 2.093
The null and alternate hypotheses is
H0 : ud= 0.5 against Ha: ud≠ 0.5 (two tailed test)
d`= ∑di/n= -7/20= 0.35
Sd²= ∑(di-d`)²/n-1 = 1/n-1 [∑di²- (∑di)²n]
= 1/19[ 887-(0.35)²/20] = [887-0.1225/19]= 46.67
Sd= 6.83
Therefore
t= d`/ sd/√n
t= 0.35/ 6.83/√20
t= 0.05124/ 4.472=0.01145
Since the calculated value of t = 0.01145 does not fall in the critical region ║t║≥ t (0.025,19) = 2.093 so we accept H0 the data suggests that after training people can correctly predict coin toss outcomes better than the 50 percent expected by chance guessing alone.
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is the answer