Answer:
The margin of error for the 99% confidence level for this sample is ±2.23.
None of the given figures is close to the answer:
E. None of the above
Step-by-step explanation:
margin of error (ME) around the mean can be calculated using the formula
ME=
where
- z is the corresponding statistic of the 99% confidence level (2.576)
- s is the population IQ standard deviation (15)
- N is the sample size (300)
Using these numbers we get:
ME=
≈ 2.23
Answer:
yes
Step-by-step explanation:
6(1)-2>3
6-2>3
4>3
Percent cost of the total spent on
each equipment = 14.23%,fees=62.45% and transportation = 23.32%.
As given in the question,
Money spent on
Each equipment = $36
Fees = $158
Transportation = $59
Total money spent = $( 36 + 158 + 59)
= $253
Percent cost of the total spent on
Each equipment = (36/253) × 100
= 14.23%
Fees = (36/253) × 100
= 62.45%
Transportation =(36/253) × 100
= 23.32%
Therefore, percent cost of the total spent on
each equipment = 14.23%,fees=62.45% and transportation = 23.32%.
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Answer:
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
The margin of error of the interval is given by:
In this problem, we have that:
99.5% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?
This is n when M = 0.07. So
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
Percent decrease. 41%
percent change= (new-old)/old *100
= -41 % the negative means decreasing
