Do the data suggest that after training people can correctly predict coin toss outcomes better than the 50 percent expected by chance guessing alone

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.

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

65                          54                                  11                          121        

∑                                                                   -7                          887      

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.

3, nah jk it's 2

step-by-step explanation:

it is 3 cause i'm sanic

answer: huh

step-by-step explanation: