The margin of error would be larger if the sample standard deviation were smaller. Question 7 Not yet answered Points out of 1.00 Not flaggedFlag question Question text The daily revenues of a cafe near the university are approximately normally distributed. The owner recently collected a random sample of 40 daily revenues and found a 90% confidence interval for the average daily revenues in his shop is (973.993, 1026.007). He is unsatisfied by the precision of this confidence interval, however, and wishes to reduce the margin of error by a factor of 2, while retaining the same level of confidence. What sample size do you suggest he use to obtain the desired margin of error?

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".

[tex]\bar X [/tex] represent the sample mean for the sample

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n=40 represent the sample size

Solution to the problem

First we can calculate the sample mean like this:

[tex] \bar X =\frac{1026.007+973.993}{2}= 1000[/tex]

And then we can find the margin of error like this:

[tex] ME = \frac{1026.007-973.993}{2}= 26.007[/tex]

And we know that that the margin of error is given by:

[tex] ME = t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]

The degrees of freedom are given by:

[tex] df = n-1 =40-1 =39[/tex]

And we can find the critical value for this case [tex] t_{\alpha/2}=1.685[/tex]

And then we can estimate the value of s

[tex] s = \frac{ME}{t_{\alpha/2}} \sqrt{n} = \frac{26.007}{1.685} * \sqrt{40} =97.616[/tex]

The margin of error for the proportion interval is given by this formula:

[tex] ME=t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (4)

And on this case we have that ME =13.0035 and we are interested in order to find the value of n, if we solve n from equation (4) we got:

[tex]n=(\frac{t_{\alpha/2} s}{ME})^2[/tex] (5)

[tex]n=(\frac{1.685(97.616)}{13.0035})^2 =160.00 \approx 160[/tex]

So the answer for this case would be n=160 rounded up to the nearest integer

princessesther2011 princessesther2011 · Answer 2 · 2020-03-13T03:39:36+01:00

Answer:

I suggest he uses a sample size of 46

Step-by-step explanation:

90% confidence interval with a sample size of 40 is (973.993, 1026.007)

Margin of error (E) = (1026.007 - 973.993) ÷ 2 = 52.014 ÷ 2 = 26.007

sd = E√n/t

n = 40

df = n-1 = 40 - 1 = 39

confidence level = 90%

t-value corresponding to 39 degrees of freedom and 90% confidence level is 1.685

sd = 26.007√40/1.685 = 97.62

Reducing E by a factor of 2 (E becomes 24 approximately) would lead to an increase in the sample size because the relationship between E and sample size is inverse in which decrease in one quantity (E) would lead to a corresponding increase in the other quantity (sample size)

If n = 46

df = n-1 = 46-1 = 45

t-value = 1.6795

E = t×sd/√n = 1.6795×97.62/√46 = 24 (to the nearest integer)