I was having trouble with 5-7 if that\'s okay. I\'m not sure how to get all of t

Question

I was having trouble with 5-7 if that's okay. I'm not sure how to get all of the pieces for the formulas.

Confidence Interval pdf X Sampling and sampling X Take Home Exam 4.pdf X Chapter 7-201757-MAT x Secure h ps://ttu.blackboard.com bbcswebdav/pid-25411 15-dt-conten rid-13090374 1 /courses MATH-2345-D01/Take %20Home%20Exam%204.pdf B) 58.2% A) 76% C) 38 D) 98.5 ll (q5-q7) A simple random sample of 100 items from a population with standard deviation of 12 resulted in a sample mean of 56. The confidence coefficient of the population mean is 0.95. 5. The level of significance is: A) 0.05 C) 0.99 D) 0.95 B) 0.01 6. The margin of error for the mean is: A) 4.52 B) 3.25 C) 2.35 D) 1.78 7. The 95% confidence interval of the mean is: A) 53.65 58.35 B) 54.24-59.38 C) 51.20-59.89 D) 54.78-56.34 Ill(q8-q9). In a survey, the planning value for the population proportion is p 0.25. 8. How large a sample should be taken to provide a 95% confidence interval with a margin of error of 0.1? B) 1200 C) 300 A) 50 D) 72 9. If the past data are not available for developing a planning value for p what value would you choose? B) 0.5 C) 0.01 D) 0.75 10. Which is the t value in the upper tail area of 0.025 with 6 degrees of freedom? C) 2.447 A 943 D) 2.015 B) 43 Ask me anything a wa 12:23 AM 3/2/2017

Explanation / Answer

Q5.
0.05

Q6.
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)=56
Standard deviation( sd )=12
Sample Size(n)=100
Margin of Error = Z a/2 * 12/ Sqrt ( 100)
= 1.96 * (1.2)
= 2.35

Q7.
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)=56
Standard deviation( sd )=12
Sample Size(n)=100
Confidence Interval = [ 56 ± Z a/2 ( 12/ Sqrt ( 100) ) ]
= [ 56 - 1.96 * (1.2) , 56 + 1.96 * (1.2) ]
= [ 53.65,58.35 ]

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