Question

you extract a sample of 115 students from this university. The sample proportion is the proportion of students in this sample who live on campus. The standard deviation of the sampling distribution of this sample proportion, rounded to four decimal places, is: 5. A random sample of 82 customers, who visited a department store, spent an average of $71 at this store. Suppose the standard deviation ofexpenditures at this store is σ. $19. The 98% confidence interval for the population mean (rounded to four decimal places) is: Answer Lower limit: Upper limit: 6. A sample of 20 elements produced a mean of 91.4 and a standard deviation of 11.16. Assuming that the population has a normal distribution, the 90% confidence interval for the population mean is: Lower limit Answer: Upper limit: 7. In a hypothesis test with hypotheses Ho : μ = 60 and H: μ关60, a random sample of 37 elements selected from the population produced a mean of 62.5. Assuming that σ-52 and the population is normally distributed, what is the approximate p-value for this test? round your answer to four decimal places)

Answer 1

(a) The summary stati stics The sample size is n 82. The sample mean isX 71 The population standard deviati on is known and it is denoted by the greek letter is The level of significance is α-002 The critical value using Excel2010 The critical value is obtained by using the Excel2010 function normsinv (probability) normsinv (0.01) 2.32635 cał 2 002.3264 (b) The calculation of the Margin of Error The formula for calculating the margin of error is The margin of error isE-z 2 Z0022 19 2001 19 2.3264 9.055385 2.3264(2.098199) 4.88125 E4.88125 Therefore, the margin of error is approximately E 488125 (c) The 98% confidence interval The formula for calculating the 98% confidence interval for the population mean is of the population mean (μ 71-4.88125S 71+4.88125 66, 1 1875 μ 75.88 125 -6 11355 813 Interpretat on We can be 98% confident that the true population mean is between 66.1188 and 75.8813