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.

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".  

represent the sample mean  

population mean (variable of interest)  

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:  

(1)  

The point estimate of the population mean is

Since the Confidence is 0.98 or 98%, the value of and , 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  

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

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

0.84 dang lotta people are looking for this lol or if you round up its 1 but the overall answer to that is 0.84

step-by-step explanation: