VA Chapter 7.1 & 7.2 HW x C Chegg Study IGuided Sc x C O www.webassign ne Respon

Question

VA Chapter 7.1 & 7.2 HW x C Chegg Study IGuided Sc x C O www.webassign ne Response /submit? dep 16015990 t/web/Student/Assignme TC 49 points I Previous Answers BBUnderstat10 7.1.019 My Notes Ask Your Teacher Thirty-two small communities in Connecticut (population near 10,000 each) gave an average of x 138.5 reported cases of larceny per year. Assume that o is known to be 43.3 cases per year. (a) Find a 90% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit 137.8 upper limit 139.2 margin of error 0.7 (b) Find a 95% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit 137.7 upper limit 139.3 margin of error 0.8 (c) Find a 99% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit 137.4 upper limit 139.6 margin of error 1.1 (d) Compare the margins of error for parts (a) through (c). As the confidence levels increase, do the margins of error increase? As the confidence level increases, the margin of error increases. O As the confidence level increases, the margin of error decreases. O As the confidence level increases, the margin of error remains the same. 4:37 PM Ask me anything 4/16/2017

Explanation / Answer

Q1.
a.
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=138.5
Standard deviation( sd )=43.3
Sample Size(n)=32
Margin of Error = Z a/2 * 43.3/ Sqrt ( 32)
= 1.64 * (7.65)
= 12.55
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=138.5
Standard deviation( sd )=43.3
Sample Size(n)=32
Confidence Interval = [ 138.5 ± Z a/2 ( 43.3/ Sqrt ( 32) ) ]
= [ 138.5 - 1.64 * (7.65) , 138.5 + 1.64 * (7.65) ]
= [ 125.95,151.05 ]

b.
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=138.5
Standard deviation( sd )=43.3
Sample Size(n)=32
Margin of Error = Z a/2 * 43.3/ Sqrt ( 32)
= 1.96 * (7.65)
= 15
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=138.5
Standard deviation( sd )=43.3
Sample Size(n)=32
Confidence Interval = [ 138.5 ± Z a/2 ( 43.3/ Sqrt ( 32) ) ]
= [ 138.5 - 1.96 * (7.65) , 138.5 + 1.96 * (7.65) ]
= [ 123.5,153.5 ]

c.
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=138.5
Standard deviation( sd )=43.3
Sample Size(n)=32
Margin of Error = Z a/2 * 43.3/ Sqrt ( 32)
= 2.58 * (7.65)
= 19.75
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=138.5
Standard deviation( sd )=43.3
Sample Size(n)=32
Confidence Interval = [ 138.5 ± Z a/2 ( 43.3/ Sqrt ( 32) ) ]
= [ 138.5 - 2.58 * (7.65) , 138.5 + 2.58 * (7.65) ]
= [ 118.75,158.25 ]

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.