answer:a. To construct a 90% confidence interval estimate for the population mean of absences per tutorial in the past 5 weeks, we can use the formula: Confidence interval = sample mean ± (critical value * standard deviation / √sample size) The critical value can be obtained from the t-distribution table, corresponding to a 90% confidence level with 24 degrees of freedom (sample size minus 1). Let's assume the critical value is 1.711 (hypothetical value for demonstration purposes). Sample mean = 11.6 absences Sample standard deviation = 4.1 absences Sample size = 25 tutorials Critical value (hypothetical) = 1.711 Confidence interval = 11.6 ± (1.711 * 4.1 / √25) Calculating the confidence interval: Confidence interval = 11.6 ± (1.711 * 4.1 / 5) Confidence interval = 11.6 ± (1.3916) Confidence interval = (10.2084, 12.9916) Therefore, with a 90% confidence level, we can estimate that the population mean of absences per tutorial in the past 5 weeks lies between 10.2084 and 12.9916. b. No, the lecturer cannot claim that on average there are 15 absences per tutorial because the value of 15 falls outside the confidence interval calculated in part a. The confidence interval suggests that the population mean lies between 10.2084 and 12.9916. c. If the lecturer feels her confidence interval is too narrow, she can either increase the sample size or increase the critical value (confidence level). By increasing the sample size, the variability in the sample mean decreases, resulting in a narrower confidence interval. Alternatively, by increasing the critical value (confidence level), the margin of error in the confidence interval will widen, leading to a wider interval.

answer:a. To calculate the sample size needed to be 95% confident of being correct to within 4 minutes, we can use the formula: Sample size = (Z * standard deviation / margin of error)^2 The Z-value corresponds to the desired confidence level (95%) and can be obtained from the standard normal distribution table. Let's assume the Z-value is 1.96 (hypothetical value for demonstration purposes). Standard deviation = 50 minutes Margin of error = 4 minutes Z-value (hypothetical) = 1.96 Sample size = (1.96 * 50 / 4)^2 Sample size = (98 / 4)^2 Sample size = 24.5^2 Sample size ≈ 600.25 Therefore, a sample size of approximately 601 listeners is needed to be 95% confident of being correct to within 4 minutes. b. If 99% confidence is desired, we need to calculate the sample size using the same formula but with a different Z-value. Let's assume the Z-value for 99% confidence is 2.58 (hypothetical value for demonstration purposes). Standard deviation = 50 minutes Margin of error = 4 minutes Z-value (hypothetical) = 2.58 Sample size = (2.58 * 50 / 4)^2 Sample size = (129 / 4)^2 Sample size = 32.25^2 Sample size ≈ 1040.0625 Therefore, a sample size of approximately 1041 listeners is needed to be 99% confident of being correct to within 4 minutes.

answer:a. To construct a 95% confidence interval estimate for the population mean frontline employee turnover cost of member organizations, we can use the formula: Confidence interval = sample mean ± (critical value * standard deviation / √sample size) Sample mean (X) = 12,500 Sample standard deviation (S) = 1,000 Sample size (n) = 100 The critical value can be obtained from the t-distribution table, corresponding to a 95% confidence level with 99 degrees of freedom (sample size minus 1). Let's assume the critical value is 1.984 (hypothetical value for demonstration purposes). Confidence interval = 12,500 ± (1.984 * 1,000 / √100) Calculating the confidence interval: Confidence interval = 12,500 ± (1.984 * 1,000 / 10) Confidence interval = 12,500 ± (198.4) Confidence interval = (12,301.6, 12,698.4) Therefore, with a 95% confidence level, we can estimate that the population mean frontline employee turnover cost of member organizations lies between 12,301.6 and 12,698.4. b. To construct a 95% confidence interval estimate for the population proportion of member organizations that have both talent and development programs in place, we can use the formula: Confidence interval = sample proportion ± (critical value * √(sample proportion * (1 - sample proportion) / sample size) Sample proportion = 30/100 = 0.3 Sample size = 100 The critical value for a 95% confidence level can be obtained from the standard normal distribution table. Let's assume the critical value is 1.96 (hypothetical value for demonstration purposes). Confidence interval = 0.3 ± (1.96 * √(0.3 * (1 - 0.3) / 100) Calculating the confidence interval: Confidence interval = 0.3 ± (1.96 * √(0.3 * 0.7 / 100)) Confidence interval = 0.3 ± (1.96 * 0.0485) Confidence interval = (0.2052, 0.3948) Therefore, with a 95% confidence level, we can estimate that the population proportion of member organizations with both talent and development programs in place lies between 0.2052 and 0.3948. c. To calculate the sample size needed to have 99% confidence of estimating the population mean frontline employee turnover cost to within 250, we can use the formula: Sample size = (Z * standard deviation / margin of error)^2 The Z-value corresponds to the desired confidence level (99%) and can be obtained from the standard normal distribution table. Let's assume the Z-value is 2.576 (hypothetical value for demonstration purposes). Standard deviation = 1,000 Margin of error = 250 Z-value (hypothetical) = 2.576 Sample size = (2.576 * 1,000 / 250)^2 Sample size = (2.576 * 4)^2 Sample size = 10.304^2 Sample size ≈ 106.128 Therefore, a sample size of approximately 107 member organizations is needed to have 99% confidence of estimating the population mean frontline employee turnover cost to within 250. d. To calculate the sample size needed to have 90% confidence of estimating the population proportion of organizations that have both talent and development programs in place to within 0.045, we can use the formula: Sample size = (Z^2 * p * (1 - p)) / E^2 The Z-value corresponds to the desired confidence level (90%) and can be obtained from the standard normal distribution table. Let's assume the Z-value is 1.645 (hypothetical value for demonstration purposes). p = proportion of member organizations with both talent and development programs in place (unknown) E = 0.045 Sample size = (1.645^2 * p * (1 - p)) / 0.045^2 Since the value of p is unknown, we cannot determine the sample size without any prior information about the population proportion. Therefore, we cannot determine the sample size needed to have 90% confidence of estimating the population proportion without additional information.

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