Question

1. Use the confidence interval limits of (0.463, 0.754) to find the point estimate (p hat) and the margin of error E.
2. A retailer would like to estimate the proportion of their customers who bought an item after viewing their website on a certain day. Using a 95% confidence level and 5% margin of error, how many customers do they have to monitor? A separate study determined 75% of customers bought an item after viewing a website. Round your answer to the nearest whole number.
3. Pertaining to question #5, assume the retailer had no prior knowledge of the separate study. Calculate the new necessary sample size, rounding your answer to the nearest whole number.
4. What is the critical value  for constructing a 95% confidence interval for a mean from a sample size of n = 16 observations?
5. Use the birth weights of baby lambs given below, in lbs, to construct a 95% confidence interval estimate of the mean birth weight of all lambs. Assume that sigma is known to be 1.5 lbs.

7.5	9.3	11.8	6.8	7	8.2
8	9.8	10	7.5	8.1	7.7
9	9.1	6.2	7.6	8.3	10.1
12	6.9	8.5	7.1	10.6	6.5
8.3	8	9.4	5.9	11.2	9

Answer 1

Q1:

Confidence interval = (0.463, 0.754)

Point estimate (p hat) = (0.463 + 0.754)/2 = 0.6085

Margin of error, E = Point estimate - lower limit = 0.6085 - 0.463 = 0.1455

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Q2:

Proportion, p = 0.75      

Margin of error, E = 0.05      

Confidence Level, CL = 0.95      

Significance level, α = 1 - CL = 0.05      

Critical value, z = NORM.S.INV(0.05/2) = 1.96

Sample size, n = (z² * p * (1-p)) / E² = (1.96² * 0.75 * 0.25)/ 0.05²          

= 288.1094 = 288

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Q3: Let Proportion, p = 0.5      

Margin of error, E = 0.05      

Confidence Level, CL = 0.95      

Significance level, α = 1 - CL = 0.05      

Critical value, z = NORM.S.INV(0.05/2) = 1.96

Sample size, n = (z² * p * (1-p)) / E² = (1.96² * 0.5 * 0.5)/ 0.05²          

= 384.1459 = 384

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Q4:

α = 1-0.95 = 0.05

df = n-1 = 15,

Critical value, t-crit = T.INV.2T(0.05, 15) = 2.131

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Q5:

∑x = 255.4  

n = 30  

Mean, x̅ = Ʃx/n = 255.4/30 = 8.5133

σ = 1.5

95% Confidence interval :  

At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) = 1.960

Lower Bound = x̅ - z_c*σ/√n = 8.5133 - 1.96 * 1.5/√30 = 7.98

Upper Bound = x̅ + z_c*σ/√n = 8.5133 + 1.96 * 1.5/√30 = 9.05

7.98 < µ < 9.05