Question

Assume that the one-way commute time of an UoU student from his house to school is a normally distributed random variable which we will call X. Furthermore, assume that the population standard deviation of X is σ = 10 minutes. Let μ be the unknown population mean for X:
Experimental design: Determine a minimum sample size such that we will be 95 % confident that the error will not exceed 5 minutes when the sample average x is used to estimate μ. Let n denote this sample size

Answer 1

Answer:

$n=(\frac{1.960(10)}{5})^2 =15.36 \approx 16$

So the answer for this case would be n=16 rounded up to the nearest integer

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".

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

$\mu$ population mean (variable of interest)

$\sigma=10$ represent the sample standard deviation

n represent the sample size  

ME=5 represent the margin of error

Solution to the problem

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

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

The margin of error is given by this formula:

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

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (3)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.96$, replacing into formula (3) we got:

$n=(\frac{1.960(10)}{5})^2 =15.36 \approx 16$

So the answer for this case would be n=16 rounded up to the nearest integer