Students in a representative sample of 65 first-year students selected from a large university in England participated in a study of academic procrastination. Each student in the sample completed the Tuckman Procrastination Scale, which measures procrastination tendencies. Scores on this scale can range from 16 to 64, with scores over 40 indicating higher levels of procrastination. For the 65 first-year students in this study, the mean score on the procrastination scale was 37.1 and the standard deviation was 6.45.

Construct a 95% confidence interval estimate of u, the population mean procrastination scale for second-year students at this college. (Use technology. Round your answers to three decimal places.)

So on this case the 95% confidence interval would be given by (35.502;38.698) seconds

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 for the sample  

population mean (variable of interest)  

s=6.45 represent the sample standard deviation  

n represent the sample size  

Confidence interval

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

(1)  

In order to calculate the critical value we need to find first the degrees of freedom, given by:  

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: "=-T.INV(0.025,64)".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 (35.502;38.698) seconds

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

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step-by-step explanation:

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