Question

A study finds that college students found it unpleasant to sit alone and think.1 The same article describes a second study in which college students appear to prefer receiving an electric shock to sitting in solitude. The article states that "when asked to spend 15 minutes in solitary thought, 12 of 18 men and 6 of 24 women voluntarily gave themselves at least one electric shock." Use this information to estimate the difference between men and women in the proportion preferring pain over solitude. The standard error of the estimate is 0.154.

Answer 1

Answer:

Assuming a confidence level of 95%.

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

The standard error for this case is given by: $SE= 0.154$

$(0.667-0.25) - 1.96*0.154=0.115$  

$(0.667-0.25) + 1.96*0.154=0.719$  

And the 95% confidence interval for the difference of the two proportions would be given (0.115;0.719).  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$p_A$ represent the real population proportion for brand A  

$\hat p_A =\frac{12}{18}=0.667$ represent the estimated proportion for male

$n_A=18$ is the sample size required for male

$p_B$ represent the real population proportion for female

$\hat p_B =\frac{6}{24}=0.25$ represent the estimated proportion for female

$n_B=24$ is the sample size required for female

$z$ represent the critical value for the margin of error  

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$  

Assuming a confidence level of 95%.

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

The standard error for this case is given by: $SE= 0.154$

And replacing into the confidence interval formula we got:  

$(0.667-0.25) - 1.96*0.154=0.115$  

$(0.667-0.25) + 1.96*0.154=0.719$  

And the 95% confidence interval for the difference of the two proportions would be given (0.115;0.719).