Question

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

Answer 1

Answer:

The sample size to obtain the desired margin of error is 160.

Step-by-step explanation:

The Margin of Error is given as

$MOE=z_{crit}\times\dfrac{\sigma}{\sqrt{n}}$

Rearranging this equation in terms of n gives

$n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2$

Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as

$n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n$

As n is given as 40 so the new sample size is given as

$n_2=4n\\n_2=4*40\\n_2=160$

So the sample size to obtain the desired margin of error is 160.