A famous study of children at Stanford in 1972 involved researchers giving a chi

Question

A famous study of children at Stanford in 1972 involved researchers giving a child a marshmallow and then asking the child to wait for 15 minutes with the promise of a second marshmallow if the child did not eat the first. A recent modification of this study looked at whether providing an unreliable or a reliable environment would affect the length of time the child waited for a second marshmallow. Please consult the description of this study found here: http://www.rochester.edu/news/show.php?id=4622

1. Between the reliable and unreliable groups there is also clearly a difference in the proportion of children who waited the full fifteen minutes. Construct a 95% confidence interval for the difference in proportion of children in unreliable and reliable environments who waited for the full 15 minutes. (Make sure that you pick a valid strategy, given what you know about the data)

2. Based on the confidence interval above, and at the 5% significance level, would we reject the claim that there is no difference in proportion of children who waited the full length of time between reliable and unreliable environments? Why?

Explanation / Answer

From the article , the proportion for the 2 groups can be created from thel line

Only one of the 14 children in the unreliable group waited the full 15 minutes, compared to nine children in the reliable condition.

so p1 = 1/14 = 0.071

and p2 = 9/14 = 0.64

Here the hypothesis would be

H0: P1 = P2
Ha: P1 P2

lets calculate the pooled standard deviation sing the formula

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = (0.071*14 +0.64*14)/(14+14) = 0.355

now the standard error is

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

sqrt{ 0.355* ( 1 - 0.355 ) * [ (1/14) + (1/14) ] } = 0.18

now Test statistic. The test statistic is a z-score (z) defined by the following equation.

z = (p1 - p2) / SE

so Z = (0.071 - 0.64)/0.18 = - 3.16

from the z table we see that the corresponding value is

0.000789

which is less than 0.05 , hence we reject the null hypothesis to concluded that the proportions are different for reliable and unreliable environments

The difference in the confidence interval is calculated as

(0.071 - 0.64) +- Z * SE , where z = 1.96 from the z tables

so (0.071 - 0.64) +- 1.96* 0.18

= -0.2162 , -0.9218

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