Using the z-distribution, we have that:
- For a 99% confidence level, a sample size of 127 is needed.
- For a 95% confidence level, a sample size of 74 is needed, meaning that a decrease in the confidence level decreases the needed sample size, as M and n are inverse proportional.
<h3>What is a z-distribution confidence interval?</h3>
The confidence interval is:
The margin of error is:
In which:
- is the sample mean.
- is the standard deviation for the population.
For a 99% confidence interval, , hence z is the value of Z that has a p-value of , so the critical value is z = 2.575.
The margin of error and population standard deviation are:
Hence we have to solve for n to find the needed sample size, as follows:
n = 126.4.
Rounding up, for a 99% confidence level, a sample size of 127 is needed.
For the 95% confidence interval, we have that z = 1.96, hence:
n = 73.3.
Rounding up, for a 95% confidence level, a sample size of 74 is needed, meaning that a decrease in the confidence level decreases the needed sample size, as M and n are inverse proportional.
More can be learned about the z-distribution at brainly.com/question/25890103
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Answer:$56.04
Step-by-step explanation: just add them all
Answer: = $33100
Step-by-step explanation:
Amount of bad debt expense is given by;
= Bad debt estimate - allowance for doubtful accounts
= 5% of account receivables - allowance for doubtful accounts
= 5/100 × 1160000 - 24900
=58000 - 24900
= $33100
Goodluck...
The question is wrong since population is not counted in points . for question like these use formula :P(1+R/100)^t where p is current population,r is rate and t is time.
I’m here to help! ( pssss, the other guy’s answer was trash) Ok:
1) 10+9= 19
2) 8+5= 13
3) 7+2=9
4) 3+9=12
5) 4+1=5
6) 2+4=6
Hope this helps!
~ Your Cookie Monster