The Confidence interval for 95% who believes that the local businesses overcharge = (0.7026, 0.7574)
<h3>
What is meant by confidence interval?</h3>
The range of values we see in our sample and hope to identify the value that most closely represents the population are referred to as a confidence interval.
<h3>According to the given information:</h3>
Sample size n = 1000
73% of town residents believed that local businesses overcharged for their products over 1000 resident.
= (1000/100) x 73
= 730
Sample proportion p = 730/1000
= .73
q = 1-p = 0.23
Std error of proportion = √(pq/n)
= √((.73*0.27)/1000)
= 0.0140
95% Z critical value = 1.96
Margin of error = 1.96*0.0140
= 0.0274
Confidence interval = sample proportion ±margin of error
(0.7026, 0.7574)
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I believe it would be A or E
After n years the amount = 1500(1 + (0.0475/4)^4n
The probable number of prople sent to US emergency rooms by 2090 can be between 22,050 and 23,100
Step-by-step explanation:
Total sent to US emergency room by 2010= 21000
The estimated increase in the rise of cases = 5 to 10% by 2090
Final numbers in 2090
Hence the final numbers in 2090 would be 5 to 10% more than the total cases in 2010
Lower limit= 5% of 21000= 1050
Hence lower limit of cases in 2090= 21000+1050= 22050
Upper limit of cases in 2090= 10% of 21000= 21000+2100= 23,100
The number would lie anywhere between 22050 and 23,100 in 2090
Answer:
Step-by-step explanation:
