Question

If amy cuddy and her research team originally established a sample size of n=200 (100 in each group) and then discovered that 86% of the high-power group (86 of 100) took a gambling risk while 60% of the low-power group (60 of 100) took the risk, then the p-value would have been much lower than that originally found in question 1. this p-value would have been less sensitive and would have provided stronger evidence for the researcher's hypothesis. state the p-value, rounding to 5 decimal places.

Answer 1

The p-value that would be given as the evidence would be given as p-value = 0.00002 and also P value = 0.00003

How to solve for the p value

The sample size of the first sample = 100

x1 = success 0f 86%

We have to find the proportion of this = 0.86

The sample size of the second sample n2 = 100

x12 = success 0f 60%

We have to find the proportion of this = 60/100

= 0.6

The difference = 0.86-0.6= 0.26

Next we have to find the pooled proportion. This is taken given as p = (x1+x2)/(n1+n2) this gives us 0.73

Next we have to find the standard error

= √(p*(1-p)*(1/n1+ 1/n2)= 0.06279

Z stat = (0.26-0)/0.0628= 4.1411

From here the p value would have to be

p-value = 0.00002 and also

P value = 0.00003

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