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allochka39001 [22]
1 year ago
12

Suppose that the researchers wanted to estimate the mean reaction time to within 6 msec with 95% confidence. Using the sample st

andard 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.)
Mathematics
1 answer:
Lisa [10]1 year ago
3 0

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  

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