Question

Researchers continue to find evidence that brains of adolescents behave quite differently than either brains of adults or brains of children. In particular, adolescents seem to hold on more strongly to fear associations than either children or adults, suggesting that frightening connections made during the teen years are particularly hard to unlearn. In one study,1 participants first learned to associate fear with a particular sound. In the second part of the study, participants heard the sound without the fear-causing mechanism, and their ability to "unlearn" the connection was measured. A physiological measure of fear was used, and larger numbers indicate less fear. We are estimating the difference in mean response between adults and teenagers. The mean response for adults in the study was 0.225 and the mean response for teenagers in the study was 0.059. We are told that the standard error of the estimate is 0.091. Let group 1 be adults and group 2 be teenagers.
(a) Give notation for the quantity that is being estimated.

Answer 1

Answer:

a) $\mu_1 -\mu_2$ parameter of interest.

Where $\mu_1$ represent the mean response for adults

$\mu_2$ represent the mean response for teenegers

b) The best estimate is given by $\bar X_1 -\bar X_2$

Since the best estimator for the true mean is the sample mean $\hat \mu = \bar X$

c) The best estimate is given by $\bar X_1 -\bar X_2 =0.225-0.059=0.166$

d) The 95% confidence interval would be given by $-0.012 \leq \mu_1 -\mu_2 \leq 0.344$  

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".  

Let group 1 be adults and group 2 be teenagers.

$\bar X_1 =0.225$ represent the sample mean 1

$\bar X_2 =0.059$ represent the sample mean 2

n1 represent the sample 1 size  

n2 represent the sample 2 size  

$s_1$ sample standard deviation for sample 1

$s_2$ sample standard deviation for sample 2

SE =0.091 represent the standard error for the estimate

(a) Give notation for the quantity that is being estimated.

$\mu_1 -\mu_2$ parameter of interest.

(b) Give notation for the quantity that gives the best estimate.

$\mu_1 -\mu_2$ parameter of interest.

The best estimate is given by $\bar X_1 -\bar X_2$

Since the best estimator for the true mean is the sample mean $\hat \mu = \bar X$

(c) Give the value for the quantity that gives the best estimate.

The best estimate is given by $\bar X_1 -\bar X_2 =0.225-0.059=0.166$

(d) Give a confidence interval for the quantity being estimated. Assuming 95% of confidence

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)  

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =0.225-0.059=0.166$

We can assume that since we know the standard error the deviations are known and we can use the z distribution instead of the t distribution for the confidence interval.

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$  

The standard error is given by the following formula:

$SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=0.091$

Given by the problem

Now we have everything in order to replace into formula (1):  

$0.166-1.96(0.091)=-0.012$  

$0.166+1.96(0.091)=0.344$  

So on this case the 95% confidence interval would be given by $-0.012 \leq \mu_1 -\mu_2 \leq 0.344$