The owner of a currency exchange bureau on an airport wishes to estimate the mean value of cash necessary for theexchange of PLN and EUR. Experience tells the owner that the demand for euro is distributed normally, with a standarddeviation equal to 4 (thousand PLN). Having observed the demand for 10 days, the owner obtained a mean of 24.4 (thousand PLN). Calculate the realization of the confidence interval for the expected demand with a confidence level of0.90.

Solution- To calculate the confidence interval for the mean demand for euros with a confidence level…

Answered: The owner of a currency exchange bureau…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Find the 95% z-interval or t-interval for the indicated parameter. (a) u x= 152, s = 31, n= 35 (b) p p= 0.4, n =71 (a) The 95% confidence interval for µ is | to (Round to two decimal places as needed.)
If confidence level is 92% Find level of signifcance alpha
A manufacturer of cell phone batteries wants to estimate the useful life of it'sthe battery (in thousand of hours.) The estimate is the be within 0.10 (100) hours.Assume a 95 percent level of confidence and that the standard deviation of theuseful life of the battery is 0.90 (900 hours). Determine the required sample size.
Consider the estimated equation from Example 4.3, which can be used to study the effects of skipping class on college GPA: 1.39 + .412 hsGPA + .015 ACT – 083 skipped (.33) (.094) n = 141, R = .234. colGPA (.026) (.011) Using the standard normal approximation, find the 95% confidence interval for BraGPA: |(ii) Can you reject the hypothesis H,: (iii) Can you reject the hypothesis Hg: BisGPA = 1 against the two-sided alternative at the 5% level? (i) 4 against the two-sided alternative at the 5% level?
The higher the confidence level is, 2 points the more precise the confidence interval is True False
The average price of 350 cellphones is Php 20,000 with a sample standard deviation of Php 2500. Find the 99% confidence interval of the true average price of the cellphones.
Find the confidence interval for estimating the population proportion for parts (a) through (c). a. 90% confidence level; n = 500; p=0.30 b. 95% confidence level; n = 220; p=0.04 c. α = 0.07; n = 345; p=0.50 Click the icon to view the standard normal table of the cumulative distribution function. a. The confidence interval is
Consider the estimated equation from Example 4.3, which can be used to study the effects of skipping class on college GPA: colGPA = 1.39.412 hsGPA+.015 ACT- .083 skipped (.33) (.094) (.011) (.026) n = 141, R² = .234. (ii) (i) Using the standard normal approximation, find the 95% confidence interval for BhsGPA- Can you reject the hypothesis Ho: ßhsGPA .4 against the two-sided alternative at the 5% level? Can you reject the hypothesis Ho: PhsGPA = 1 against the two-sided alternative at the 5% level? - (iii)
Find mean 2,3,53,63,75,61,53
2. The average hour spent in computer by 20 senior high school students during online class is 7 hours with a standard deviation of 2 hours. Construct a 98% confidence interval of the average hour spent by all senior high school students. Identify the width of the interval. 20
Find the lower bound of the confidence interval, rounding the result to four decimal places. Р(1 — р) p - E = P – Z«j2\ %3D 0.57(1 – 0.57) = 0.57 1.960 1,007 Find the upper bound of the confidence interval, rounding the result to four decimal places. P(1 — р) p + E : p + z 0.57(1 – 0.57) = 0.57 + 1.960 1,007 Therefore, a 95% confidence interval for the population proportion of adults who lack confidence they will be able to afford health insurance in the future is from a lower bound of X to an upper bound of
Question of the day. The mean weekly sales of the chocolate bar in candy stores was 146-3 bars per store. After an advertising campaign the mean weekly sales in 22 stores for a typical week increased to 153-7 and showed a standard deviation of 17-2. Was the advertising campaign successful?

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