Asurvey of 361 students recorded the number of hours of television they watched per week. the sample mean was 6.504 hours with a standard deviation of 5.584. the standard error of the mean was 0.294. find a 90% confidence interval for the population mean.

The 90% confidence interval would be given by (6.022;6.986)    

Step-by-step explanation:

1) 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 for the sample  

population mean (variable of interest)

s=5.584 represent the sample standard deviation

n=361 represent the sample size  

2) Confidence interval

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

  (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

(2)  

(3)  

The mean calculated for this case is

The sample deviation calculated

Since the sample size is large enough >30 we can use the z distirbution as approximation of the t distribution.

Since the Confidence is 0.90 or 90%, the value of and , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that

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

So on this case the 90% confidence interval would be given by (6.022;6.986)    

where is the diagram? shown below where? ?

step-by-step explanation:

it is 10.6 according to my calculations