Asurvey shows that college students spend on average about $20 per month on yum good ramen. a. assume that the survey was conducted on 200 students and assume a standard deviation of $6. find the 95% confidence interval of the estimate. b. assume that the survey was conducted on 200 students and assume a standard deviation of $3. find the 95% confidence interval of the estimate. c. assume that the survey was conducted on 400 students and assume a standard deviation of $6. find the 95% confidence interval of the estimate.

a) The 95% confidence interval would be given by (19.168;20.832)

b) The 95% confidence interval would be given by (19.182;20.416)

c) The 95% confidence interval would be given by (19.412;20.588)

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 represent the sample size  

Part a

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

  (1)

Since the Confidence is 0.95 or 95%, 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(0.025,0,1)".And we see that

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

So on this case the 95% confidence interval would be given by (19.168;20.832)

Part b

For this case we just need to change the value for the deviation

So on this case the 95% confidence interval would be given by (19.182;20.416)

Part c

For this case we change again the deviation and the sample size.

So on this case the 95% confidence interval would be given by (19.412;20.588)

n

step-by-step explanation:

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step-by-step explanation: