Question

15 marks) 1. Complete this question by hand and making use of appropriate tables, round results to 2 decimal places. The administrators for a hospital wish to estimate the time to register a new patient using their new software. A random sample of 15 patients produced a mean of = 11.4 minutes. The population is assumed to be approximately normal. a) Using the given sample data, construct a symmetric 80% confidence interval for the true mean time to register a new patient, assuming: The standard deviation of the population is a = 4.3 minutes. Include a concluding statement The population standard deviation is unknown and the sample standard deviation of s = 4.3 minutes. Include a concluding statement b) Compare the two confidence intervals from part (a).

Answer 1

The solution is done showing all steps with calculations and required explanations. Hope you can understand and appreciate the work.

Random sample size = n = 15 Mean .: 11.4 a) Confidence Level = 80% 3) Population Standard deviation = = 413 So, we will zo score to find confidence interval. :. Corresponding 2 - Score = 1.282 - Z Confidence Interval for true mean time (M) is : 1 - 2 5 ha ł - z o amant z o da & 1104-11282 4.3 VIS < M 11.4 +1.282 4.3 We are 9.98 M2 12.82 (Aus i 80% confident that the true population mean will lie in this confidence interval, ii) Here, population standard deviation is unknown, we will to find confidence interval. degrees of freedom = df = n- sit-scovelty of df = 14 & confidence level = 80%=1.345 Confidence Interval for true mean time () is ă-te a ama att t-slove so use I am 4.3 Vis <MT 11.4+ 1.345 4.3 15 on 11.4 - 1.345 (Aus ii) 9.91 < M < 12.89 are We 80% confident that the true population mean will lie in this confidence interval, can b) from part (a); we see that the confidence interval computed in i is wider than confidence interval in (i). This is because the t-scores are greater than a-scores for fixed soumple size and confidence level.