Show All Your Work And Formulas Required
You Are Required To Show All Of Your Workformulas71 Given A Normal D
You are required to show all of your work/formulas 7.1 Given a normal distribution with m=100 and s=10, if you select a sample of n=25, what is the probability that X is a. less than 95? b. between 95 and 97.5? c. above 102.2? d. There is a 65% chance that X is above what value? 7.17 The Chartered Financial Analyst (CFA) Institute reported that 49% of its U.S. members indicate that lack of ethical culture within financial firms has contributed the most to the lack of trust in the financial industry. (Source: “Global Market Sentiment Survey 2014,†cfa.is/Psk5O3.) Suppose that you select a sample of 100 CFA members. a. What is the probability that the sample percentage indicating that lack of ethical culture within financial firms has contributed the most to the lack of trust in the financial industry will be between 48% and 53%? b. The probability is 90% that the sample percentage will be contained within what symmetrical limits of the population percentage? c. The probability is 95% that the sample percentage will be contained within what symmetrical limits of the population percentage? d. Suppose you selected a sample of 400 CFA members. How does this change your answers in (a) through (c)? 8.1 If X=85, s=8, and n=64, construct a 95% confidence interval estimate for the population mean, m. 8.9 A bottled water distributor wants to estimate the amount of water contained in 1-gallon bottles purchased from a nationally known water bottling company. The water bottling company’s specifications state that the standard deviation of the amount of water is equal to 0.02 gallon. A random sample of 50 bottles is selected, and the sample mean amount of water per 1-gallon bottle is 0.995 gallon a. Construct a 99% confidence interval estimate for the population mean amount of water included in a 1-gallon bottle. b. On the basis of these results, do you think that the distributor has a right to complain to the water bottling company? Why? c. Must you assume that the population amount of water per bottle is normally distributed here? Explain. d. Construct a 95% confidence interval estimate. How does this change your answer to (b)? 8.11 If X=75, S=24, and n=36, and assuming that the population is normally distributed, construct a 95% confidence interval estimate for the population mean, m. 8.17 The U.S. Department of Transportation requires tire manufacturers to provide tire performance information on the sidewall of a tire to better inform prospective customers as they make purchasing decisions. One very important measure of tire performance is the tread wear index, which indicates the tire’s resistance to tread wear compared with a tire graded with a base of 100. A tire with a grade of 200 should last twice as long, on average, as a tire graded with a base of 100. A consumer organization wants to estimate the actual tread wear index of a brand name of tires that claims “graded 200†on the sidewall of the tire. A random sample of n=18 indicates a sample mean tread wear index of 195.3 and a sample standard deviation of 21.4. a. Assuming that the population of tread wear indexes is normally distributed, construct a 95% confidence interval estimate for the population mean tread wear index for tires produced by this manufacturer under this brand name. b. Do you think that the consumer organization should accuse the manufacturer of producing tires that do not meet the performance information provided on the sidewall of the tire? Explain. c. Explain why an observed tread wear index of 210 for a particular tire is not unusual, even though it is outside the confidence interval developed in (a). 8.33 What are the global trends that technology CEOs believe will transform their business? According to a PwC white paper, 105 of 117 technology CEOs from around the world responded that technological advances will transform their business and 42 responded that resource scarcity and climate change will transform their business. (Source: Fit for the Future: 17th Annual Global CEO Survey, available at pwc.to/PRQZYr.) a. Construct a 95% confidence interval estimate for the population proportion of tech CEOs who indicate technological advances as one of the global trends that will transform their business. b. Construct a 95% confidence interval estimate for the population proportion of tech CEOs who indicate resource scarcity and climate change as one of the global trends that will transform their business. c. Interpret the intervals in (a) and (b). 8.37 If you want to be 95% confident of estimating the population proportion to within a sampling error of {0.02 and there is historical evidence that the population proportion is approximately 0.40, what sample size is needed? 8.41 If the inspection division of a county weights and measures department wants to estimate the mean amount of soft-drink fill in 2-liter bottles to within {0.01 liter with 95% confidence and also assumes that the standard deviation is 0.05 liter, what sample size is needed?
Paper For Above instruction
This comprehensive statistical analysis addresses a variety of problems related to normal distributions, confidence intervals, hypothesis testing, and proportions. Each problem is approached with precise calculations and detailed explanations to elucidate the underlying statistical principles and practical implications.
Problem 1: Probabilities in Normal Distribution
Given a normal distribution with mean (μ) = 100 and standard deviation (σ) = 10, and a sample size (n) = 25, we analyze the probabilities concerning sample means (X̄). Since the sample size is 25, the standard error (SE) of the mean is calculated as:
SE = σ / √n = 10 / √25 = 10 / 5 = 2
The sampling distribution of the mean is normal with mean 100 and standard error 2.
a. Probability that X̄
Compute the z-score:
z = (95 - μ) / SE = (95 - 100) / 2 = -5 / 2 = -2.5
Using standard normal tables or calculator, P(Z
Thus, the probability that the sample mean is less than 95 is approximately 0.62%.
b. Probability that X̄ is between 95 and 97.5:
Calculate z-scores:
z1 = (95 - 100) / 2 = -2.5
z2 = (97.5 - 100) / 2 = -1.25
P(-2.5
Approximately 9.94% probability.
c. Probability that X̄ > 102.2:
Calculate z-score:
z = (102.2 - 100) / 2 = 2.2 / 2 = 1.1
P(Z > 1.1) = 1 - P(Z
About 13.57% chance.
d. To find the value X̄ above which there is a 65% chance:
P(Z > z) = 0.65 ⇒ P(Z
From standard normal tables, z ≈ -0.39.
Calculate:
X̄ = μ + z*SE = 100 + (-0.39)(2) = 100 - 0.78 = 99.22
Therefore, there is a 65% chance that X̄ exceeds approximately 99.22.
Problem 2: Sample Proportion and Confidence Intervals
Data from CFA members indicate that 49% (p̂ = 0.49) of 100 members believe lack of ethical culture affects trust. Using the normal approximation for proportions:
a. Probability that the sample percentage is between 48% and 53%:
Standard error (SE) for proportion:
SE = √[p(1 - p) / n] = √[0.49 * 0.51 / 100] ≈ 0.0504
Calculate the z-scores for 48% and 53%:
z1 = (0.48 - 0.49) / 0.0504 ≈ -0.198
z2 = (0.53 - 0.49) / 0.0504 ≈ 0.793
Probability between these z-values: P(-0.198
Approximately 36.4% probability.
b. Symmetrical limits with 90% confidence:
Critical z-value for 90% confidence: z ≈ 1.645
Margin of error (MOE):
MOE = z SE = 1.645 0.0504 ≈ 0.083
Limits:
Lower: 0.49 - 0.083 = 0.407
Upper: 0.49 + 0.083 = 0.573
c. For 95% confidence, z ≈ 1.96:
MOE = 1.96 * 0.0504 ≈ 0.099
Limits:
Lower: 0.49 - 0.099 ≈ 0.391
Upper: 0.49 + 0.099 ≈ 0.589
d. With a sample size of 400:
New SE: √[0.49 * 0.51 / 400] ≈ 0.0252
Using same z-values:
For 90% confidence:
MOE: 1.645 * 0.0252 ≈ 0.0415
Limits: 0.49 ± 0.0415 → (0.4485, 0.5315)
For 95% confidence:
MOE: 1.96 * 0.0252 ≈ 0.0494
Limits: 0.49 ± 0.0494 → (0.4406, 0.5394)
Problem 3: Confidence Interval for a Mean
X̄ = 85, standard deviation s=8, n=64. Confidence level: 95%.
Standard error:
SE = s / √n = 8 / 8 = 1
Critical z-value for 95% confidence: 1.96.
Margin of error:
ME = 1.96 * 1 = 1.96
Confidence interval:
Lower bound: 85 - 1.96 = 83.04
Upper bound: 85 + 1.96 = 86.96
Therefore, the 95% confidence interval for the population mean is approximately (83.04, 86.96).
Problem 4: Estimating Population Mean with Sample Data
Sample mean = 0.995, standard deviation = 0.02, sample size = 50. Confidence level: 99%.
Standard error:
SE = 0.02 / √50 ≈ 0.00283
Critical z-value for 99% confidence: 2.576.
Margin of error:
ME = 2.576 * 0.00283 ≈ 0.00729
Confidence interval:
Lower: 0.995 - 0.00729 ≈ 0.9877
Upper: 0.995 + 0.00729 ≈ 1.0023
This interval suggests the true mean is between approximately 0.9877 and 1.0023 gallons.
Since the specified amount is 1 gallon, and the interval slightly exceeds it, the distributor may have a valid concern.
Population normality assumption is necessary because the sample size is moderate but sufficiently large (n=50), enabling the Central Limit Theorem to justify normal approximation.
For a 95% confidence interval, z ≈ 1.96, leading to a narrower interval, which may sharpen or relax such conclusions.
Problem 5: Confidence Interval for Mean with Smaller Sample
X̄ = 75, S=24, n=36. Assuming normality,
SE = S / √n = 24 / 6 = 4
Critical t-value for 95% confidence and df=35: approximately 2.030.
Margin of error:
ME = 2.030 * 4 ≈ 8.12
Confidence interval:
Lower: 75 - 8.12 ≈ 66.88
Upper: 75 + 8.12 ≈ 83.12
The population mean is estimated within approximately (66.88, 83.12). The wider interval reflects the uncertainty given the sample size and variability.
Problem 6: Tread Wear Index Estimation
A sample of 18 tires indicates a mean tread wear index of 195.3 with a standard deviation of 21.4, claiming a population mean of 200. The confidence interval for the population mean, assuming normality, is calculated with t-distribution:
Degrees of freedom: df=17.
Critical t-value for 95% confidence: approximately 2.11.
Standard error:
SE = 21.4 / √18 ≈ 5.05
Margin of error:
ME = 2.11 * 5.05 ≈ 10.67
Interval:
Lower: 195.3 - 10.67 ≈ 184.63
Upper: 195.3 + 10.67 ≈ 205.97
This suggests the true mean tread wear index likely lies within (184.63, 205.97). Since 200 is within this range, the manufacturer’s claim appears consistent with data.
Even an observed value of 210, outside this interval, is not unusual due to natural variability and sampling error inherent in finite samples.
Problem 7: Confidence Intervals for Proportions
From responses of 105 of 117 CEOs, the estimated proportion is:
p̂ = 105/117 ≈ 0.8974.
Standard error:
SE = √[p̂(1 - p̂) / n] ≈ √[0.8974 * 0.1026 / 117] ≈ 0.0279.
a. 95% Confidence interval:
z ≈ 1.96.
MOE = 1.96 * 0.0279 ≈ 0.0547.
Interval: (0.8974 - 0.0547, 0.8974 + 0.0547) ≈ (0.843, 0.952).
b. For resource scarcity and climate change, 42 of 117 responses yield:
p̂ ≈ 0.3589, SE ≈ 0.0456.
MOE: 1.96 * 0.0456 ≈ 0.0894.
Interval: (0.2695, 0.4483).
c. These intervals indicate a high level of confidence that the true proportion of CEOs perceiving these trends falls within the respective ranges, reflecting substantial agreement on technological impact and a moderate perceived influence of resource scarcity and climate change.
Problem 8: Sample Size Calculations
To estimate the population proportion with margin of error 0.02 and known proportion 0.40, the sample size n is determined by:
n = (z / E)^2 p(1 - p) = (1.96 / 0.02)^2 0.4 * 0.6 ≈ 576.
Similarly, to estimate the mean amount of soft drink fill with error 0.01 liters, standard deviation 0.05 liters, for 95% confidence,
n = (z σ / E)^2 = (1.96 0.05 / 0.01)^2 ≈ 96.
Thus, a sample of approximately 576 units is needed for proportion estimation, and about 96 units for mean estimation, ensuring desired confidence and precision.
Conclusion
These statistical applications underline the importance of understanding sampling distributions, confidence intervals, hypothesis testing, and sample size determination