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.)

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".

[tex]\bar X=530[/tex] represent the sample mean for the sample

[tex]\mu[/tex] 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:

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

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

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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 [tex]z_{\alpha/2}=2.01[/tex]

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

[tex]530-2.01\frac{70}{\sqrt{48}}=509.692[/tex]

[tex]530+2.01\frac{70}{\sqrt{48}}=550.308[/tex]

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:

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

Assuming that [tex]\hat \sigma =s[/tex]

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:

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

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

[tex]n=(\frac{1.96(70)}{6})^2 =522.88 \approx 523[/tex]

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