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

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

b) n=523 rounded up to the nearest integer

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)(s)/(√(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(70)/(√(48))=509.692`

`530+2.01(70)/(√(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)(\sigma)/(√(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=((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=((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