Suppose that a marketing research firm wants to conduct a survey to estimate the meanμof the distribution of the amount spent on entertainment by each adult who visits a certain popularresort. the firm would like to estimate the mean of this distribution to within $60 with 95% confidence. from data regarding past operations at the resort, it has been estimated that the standard deviation ofthe entertainment expenditures is no more than $400. how large does the firm’s sample size need to be?

The firm's sample size must be of at least 171 adults.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the length of the sample.

In this problem, we have that:

. So

The firm's sample size must be of at least 171 adults.

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)