Question

Suppose that the researchers wanted to estimate the mean reaction time to within 6 msec with 95% confidence. Using the sample standard deviation from the study described as a preliminary estimate of the standard deviation of reaction times, compute the required sample size. (Round your answer up to the nearest whole number.)

Answer 1

Answer:

a) The 95% confidence interval would be given by (509.592;550.308)  

b) n=523 rounded up to the nearest integer  

Step-by-step explanation:

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".  

$\bar X=530$ represent the sample mean for the sample  

$\mu$ population mean

s=70 represent the sample standard deviation  

n=48 represent the sample size (variable of interest)  

Confidence =95% or 0.995

Part a

The confidence interval for the mean is given by the following formula:  

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)  

The degrees of freedom are df=n-1=48-1=47

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: "=-T.INV(0.025,47)".And we see that $z_{\alpha/2}=2.01$  

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

$530-2.01\frac{70}{\sqrt{48}}=509.692$  

$530+2.01\frac{70}{\sqrt{48}}=550.308$  

So on this case the 95% confidence interval would be given by (509.592;550.308)  

Part b

The margin of error is given by this formula:  

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)  

Assuming that $\hat \sigma =s$

And on this case we have that ME =6msec, and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)  

The critical value for 95% of confidence interval is provided, $z_{\alpha/2}=1.96$, replacing into formula (b) we got:  

$n=(\frac{1.96(70)}{6})^2 =522.88 \approx 523$  

So the answer for this case would be n=523 rounded up to the nearest integer