A group of local businessmen is thinking about developing land into a shopping mall. To evaluate the desirability of the location, they count the number of shoppers whovisit the neighboring shopping center each day. A random sample of 54 days reveals a daily average of 126 shoppers with a standard deviation of 48 shoppers. Thebusinessmen will develop the land if the average number of shoppers per day is more than 110. Based on the sample data, should the businessmen develop the land?Perform a hypothesis test and use a significance level of a = 0.10. Assume the population of the number of daily shoppers is approximately normally distributed.Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.

TO determine whether the local businessmen should develop the land based on the number of daily…

Answered: A group of local businessmen is…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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this population, and the mean of each sample is determined. The mean, ux,is The standard error of the mean, Oy, is (Round your answer to 2 decimal places). QUESTION 4 The average number of miles driven each day by a group of drivers is 20.7 miles with a standard deviation of 6.5 miles. If you randomly select 50 drivers, what is the probability that the average number of miles driven each day is between 19.4 and 22.5 miles? Round your answer to 4 decimal places. Click Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answers 5,337 MAY 24 MacBook Air B口 F3 888 F2 F4 1880 %23 24 4 3 T
Americans spend an average of 23.6 hours per week online, with a standard deviation of 7. The distribution of internet watching is not necessarily normally distributed. Suppose you took a random sample of 50 people. What is the probability that the sample average will be less than 23 hrs? Round your answer to three decimal places, eg 0.628.
The mean was found to be u = 70 so now we need to find the standard deviation o. The standard deviation is calculated as follows where n is the sample size, 350, and p is the probability of a success, 0.2. Find the standard deviation. V np(1 – p) O = V 350( )(1 – 0.2)
A small retail store has customers whose ages are normally distributed, with a mean of 37.5 and a standard deviation of 7.6. What is the probability that a randomly selected customer is younger than 25 years old? Round your answer to three decimal places, e.g. 0.685.
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A manufacturer claims that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor. A sample of 1010 phones from the manufacturer had a mean range of 11801180 feet with a standard deviation of 4141 feet. A sample of 2020 similar phones from its competitor had a mean range of 11101110 feet with a standard deviation of 3535 feet. Do the results support the manufacturer's claim? Let μ1�1 be the true mean range of the manufacturer's cordless telephone and μ2�2 be the true mean range of the competitor's cordless telephone. Use a significance level of α=0.01�=0.01 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 2 of 4 : Compute the value of the t test statistic. Round your answer to three decimal places.
The life of button batteries is normally distributed with a mean of 68.1 hours and a standard deviation of 18 hours 936 of the button batteries will last more than how many hours? (2 decimal places)
A health expert evaluates the sleeping patterns of adults. Each week she randomly selects 40 adults and calculates their average sleep time. Over many weeks, she finds that 5% of average sleep time is less than 3 hours and 5% of average sleep time is more than 3.4 hours. What are the mean and standard deviation (in hours) of sleep time for the population? (Round "Mean" to 1 decimal places and "standard deviation" to 3 decimal places.) Mean Standard deviation
The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return. The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is o = $2,400. (a) What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the following sample sizes: 20, 60, 100, and 300? (Round your answers to four decimal places.) sample size n = 20 sample sizen = 60 sample sizen = 100 sample size n = 300 (b) What is the advantage of a larger sample size when attempting to estimate the population mean? A larger sample has a standard error that is closer to the population standard deviation. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean. A larger sample increases the probability that the…
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Suppose a baker wants to know, on average, how many pastries they sell in a day. They sampled over 4 days. On the first day they sold 10, on the second day they sold 12, on the third day they sold 18, and on the fourth day they sold 16. The population standard deviation of pastry sales is 3. What is the point estimate (average) number of pastries sold? 26 08 056 O 14 Nex
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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