Sampling, Sampling Distributions & Estimations with Confidence
Step-by-step generalized solution summary for examples: 7.8, 8.4 – 8.8, 8.12 – 8.15, 8.19, 8.21, and 8.30Example 7.8 – Application of Conceptsa. Where the mean strength of the new concrete block surface is no different (meaning equal) from that of the original surface, this condition precisely describes the properties of the sampling distribution of the sample mean, where the sampling distribution of the sample mean is equal to the mean of the population mean from which it was selected. The sampling distribution is equal to the mean of the sampling population. b.Find the probability that sample mean and the sample mean load classification (LCN) of the 25 concrete block sections, exceeds 65. The sampling distribution is approximately normal Step 1: Compute the desired area by obtaining the z score for x? =65Answer:Z = (x??- μX)/σX = 65-602 = 2.50Step 2:Construct a probability equation based upon the question.Answer:P (x? ≥ 65) = P (z ≥ 2.50) = .5 – A (source; Figure 7.12) = .5 - .4938 = .0062c. What can be inferred about the true mean LCN of the new surface?Answer:The original assumption about the concrete blocks was incorrect. Example 8.4 – Theoretical Interpretation of a Confidence LevelQuestion:Interpret the result for the 95% Confidence Intervals for ? for 40 Random Samples of 100 sale Prices from Appendix A.1.Answer:Since the majority of the 40 repetitions of the confidence interval procedure described contain the mean of the population ($106,405), we can be assured that the statistical procedure used will result in and value that represents the same majority of the population. Example 8.5 – Finding Z for a 90% Confidence IntervalQuestion:Determine the value of Z∝/2 that would be used in constructing a 90% confidence interval of a population mean based on a large sample.SOLUTIONStep 1: Use the given value of α/2 = .05 to confirm that an area of .05 in the upper tail f the standard normal distribution and since the total area to the right of 0 is .50, z.05 is such that the area between 0 and z.05 is .050 - .05 = .45.Step 2:Find the corresponding table, find z.05 = 1.645We know that the sample mean is ± 1.645σ subscript the sample meanExample 8.6 – 99% Confidence IntervalQuestion: Estimate ?, the mean ration of sale price to total appraised value for all properties sold in 1993, using a 99% confidence interval. Interpret the interval in terms of the problem.SOLUTIONStep 1:Establish the sample meanAnswerx? ± 2.58 σnStep 2:Substitute the values for the sample mean and standard deviationAnswer:1.315 ± 2.58 ( .36650 )With a confidence level of 99% that the interval encloses the true mean ratio of sale price to total appraised value for all residential properties sold in six Tampa neighborhoods in 93’. We can also conclude that there is a general tendency for the sale prince of a property in these neighborhoods to exceed its total appraised value. This variance is likely the result of human error in the process of appraising. Example 8.7 – Effect of (1-α) on the Width of the Confidence IntervalQuestion:a. Construct a 95% confidence interval for the mean ratio of sale price to total appraised value for properties sold in the six neighborhoods.SOLUTIONStep 1:x? ± 1.96 σn ≈x? ± 1.96 snAnswer:= 1.315 ± 1.96 ( .36650 )= 1.315 ± .101 or (1.214, 1.416)Question:b. For a fixed sample size, how is the width of the confidence interval related to the confidence coefficient?SOLUTIONAnswer:The Relationship between width of confidence interval and confidence coefficient guideline informs us that as the width increases we have a greater chance that the width will contain the true parameter value. Example 8.8 – Effect of n of the Width of the Confidence IntervalQuestion:a.Construct a 99% confidence interval for ?, the population mean ratio of sale price to total appraised value.SOLUTIONStep 1:Substitute the values of the sample statistics into the general formula for a 99% confidence interval Answer:x? ± 2.58 σn ≈1.315 ± 2.58 ( .36650 ) = 1.315 ± .094 Question:b. For a fixed confidence, how is the width of the confidence interval related to the sample size?Answer:The relationship between width and confidence interval and sample size guideline informs us that the width of the confidence interval decreases as the sample size increases. Simply, larger samples provide more information than do small samples. Example 8.12 – Computer Analysis Question:Referring to example 8.6, compare the results to the interval calculated using the t score and z scores.Answer:The computed interval calculated by the MINITAB uses a t score when the population standard deviation is unknown. A calculated z score renders a less precise interval because in this case the population parameters are not well defined. This is precisely why computers use a t score. Example 8.13 – Selecting a Point EstimateQuestion:Estimate π, the true proportion of all state lottery winners (at least $50,000) who quit their jobs during the first year after striking it rich.Step 1: Calculate the sample proportion pAnswer:p = xn = Number of lottery winners in sample who quit jobTotal number of lottery winners in sampleP = 63576 = .11 (11%)Example 8.14 – 95% Confidence Interval for πQuestion: Construct a 95% confidence interval forπ, the population proportion of state lottery winners who quit their job within 1 year of striking it rich.Step 1: Substitute the values into the formula for the 95% confidence interval Answer:ρ ±zα/2 pqn = .11 ± 1.96 .11.89576= .11 ± .03Example 8.15 – 90% Confidence Interval for πQuestion:Estimate π, using a 90% confidence interval. Interpret the intervalSOLUTIONStep 1:Define the sample proportion of families that watched the premiere of “Seinfeld”Answer:p = xn = Number of families in sample that watched the premiereNumber of families in sampleP = 101165 = .612Step 2:Insert q into the 90% confidence formulaAnswer:p ± z.05 pqn = .612 ± 1.645 .612(.388)165 = .612 ± .062Example 8.19 – Small Sample 95% Confidence Interval for ?1 - ?2Questions:a. Construct a 95% confidence interval for the difference between the mean buyer savings of the two strategies. SOLUTIONStep 1:Let ?1 and ?2 represent the true mean savings of buyers using the competitive and comparative bargaining strategies.Step 2:Compute an estimate of this common varianceAnswer:Sp2 = n1-1S12+n2-1S22n1+n2-2 = 8-1(538)?+8-1( 357)?8+8-2= 208,446Step 3:Substitute the appropriate quantities into the general formula and solve Answer:(1,706 – 2,106) ± 2.145 208.446 (18+ 1 8)= -400 ± 490Example 8.21- 95% Confidence Interval for ?dQuestion:Find a 95% confidence interval for the difference in mean levels of assertiveness, ?d = (?1 - ?2).SOLUTIONStep 1Make necessary assumptions based on the sample size.Answer: Since the sample is small (n = 10) we must assume that the differences are from an approximately normal population. Step 2:Substitute the values=11.0 ± 4.7Example 8.30 – 95% Confidence Interval for σ?Question:Construct a 95% confidence interval for the true variation in tar contents of domestic cigarette brands.SOLUTIONAnswer:(n-1) s?X?a/2 ≤ σ? ≤ (n-1)s?X(1-α2)2Step 1: Substitute valuesAnswer:500-1(4.93)?563.852 ≤ σ? ≤ 500-1(4.93)?439.93621.51 ≤ σ? ≤ 27.57 ................
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