Question

Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 11 points and an unknown population mean. A random sample of 15 scores is taken and gives a sample mean of 101 points. Find the confidence interval for the population mean with a 98% confidence level.

Answer 1

Answer:

$101-2.33\frac{11}{\sqrt{15}}=94.382$  

$101+2.33\frac{11}{\sqrt{15}}=107.618$  

So on this case the 98% confidence interval would be given by (94.382;107.618)  

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".  

$\bar X=101$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$\sigma=11$ represent the population standard deviation  

n=15 represent the sample size  

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 point estimate of the population mean is $\hat \mu = \bar X =11$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that $z_{\alpha/2}=2.33$  

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

$101-2.33\frac{11}{\sqrt{15}}=94.382$  

$101+2.33\frac{11}{\sqrt{15}}=107.618$  

So on this case the 98% confidence interval would be given by (94.382;107.618)