# Suppose a random sample of 49 is selected from a population

Suppose a random sample of 49 is selected from a population with a standard deviation of 14. If the sample mean is 125, the 99% confidence interval to estimate the population mean is between ____________ .120.00 and 130.00119.85 and 133.15119.85 and 130.15119.85 and 135.15118.00 and 132.00
A random sample of 49 selected from a population with a standard deviation of 14 yielded a mean = 125. The mean and the standard deviation of the distribution of the sample means are ____________ .125 and 2125 and 49120 and 2125 and 1.33125 and 14For a random sample of 36 items and a sample mean of 211, compute a 95% confidence interval for μ if the population standard deviation is 23. Round the answers to 2 decimal places.≤ μ ≤ .The tolerance is +/­ 0.05A random sample of size 70 is taken from a population that has a variance of 49. The sample mean is 90.4. What is the point estimate of μ? Construct a 94% confidence interval for μ.*Round your answers to 2 decimal places, the tolerance is +/­0.01.The point estimate of μ is .The 94% confidence interval: * ≤ μ ≤ *Question 14Use the following information to construct the confidence intervals specified to estimate μ.a. 95% confidence for = 26, σ = 4.5, and n = 60b. 98% confidence for = 117.6, σ = 26.89, and n = 75c. 90% confidence for = 2.419, σ = 0.891, and n = 32d. 80% confidence for = 53.7, σ = 13.1, N = 500, and n = 47*Round your answer to 2 decimal places, the tolerance is +/­0.01.
**Round your answer to 3 decimal places, the tolerance is +/­0.001.a. * ≤ μ ≤ *b. * ≤ μ ≤ *c. **≤ μ ≤ **d. * ≤ μ ≤ *A candy company fills a 20­ounce package of Halloween candy with individually wrapped pieces of candy. The number of pieces of candy per package varies because the package is sold by weight. The company wants to estimate the number of pieces per package. Inspectors randomly sample 120 packages of this candy and count the number of pieces in each package. They find that the sample mean number of pieces is 18.72. Assuming a population standard deviation of .8735, what is the point estimate of the number of pieces per package? Construct a 99% confidence interval to estimate the mean number of pieces per package for the population.Round the answers to 2 decimal places.Point Estimate≤ μ ≤The tolerance is +/­ 0.05.The average total dollar purchase at a convenience store is less than that at a supermarket. Despite smaller­ticket purchases, convenience stores can still be profitable because of the size of operation, volume of business, and the markup. A researcher is interested in estimating the average purchase amount for convenience stores in suburban Long Island. To do so, she randomly sampled 24 purchases from several convenience stores in suburban Long Island and tabulated the amounts to the nearest dollar. Use the following data to construct a 90% confidence interval for the population average amount of purchases. Assume that the population standard deviation is 3.23 and the population is normally distributed.\$25144
\$11471
\$8263
\$7136
\$91078
\$3824
Round the answers to 3 decimal places.≤ μ ≤The tolerance is +/­ 0.005.In order to construct a 90% confidence interval for the population mean when the population standard deviation σ is unknown and the sample of size n = 18, the appropriate t­value to use is represented by ____________.t 0.90, 18t 0.05, 18t 0.05, 17t 0.10, 18In order to construct a 99% confidence interval for the population mean when the population standard deviation σ is unknown and the sample of size n = 20, the appropriate t­value to use is represented by ____________.t 0.01, 19t 0.05, 19t 0.005, 19
t 0.99, 20The table t value associated with the upper 5% of the t distribution and 12 degrees of freedom is __________.2.1793.3301.7823.0551.356The table t value associated with the upper 10% of the t distribution and 23 degrees of freedom is __________.1.7141.3191.3212.0692.332A national magazine marketing firm attempts towing subscribers with a mail campaign that involves a contest using magazine stickers. Often when people subscribe to magazines in this manner they sign up for multiple magazine subscriptions. Suppose the marketing firm wants to estimate the average number of subscriptions per customer of those who purchase at least one subscription. To do so, the marketing firm’s researcher randomly selects 65 returned contest entries. Twenty­seven contain subscription requests. Of the 27, the average number of subscriptions is 2.10, with a standard deviation of .86. The researcher uses this information to compute a 98% confidence interval toestimate and assumes that x is normally distributed. What does the researcher find?Round the answers to 2 decimal places.≤ μ ≤The tolerance is +/­ 0.05.According to a survey by Topaz Enterprises, a travel auditing company, the average error by travel agents is \$128. Suppose this figure was obtained from a random sample of 41 travel agents and the sample standard deviation is \$21. What is the point estimate of the national average error for all travel agents? Compute a 98% confidence interval for the national average error based on these sample results. Assume the travel agent errors are normally distributed in the population. How wide is the interval? Interpret the interval.Round the answers to 3 decimal places where necessary.Point Estimate = \$ .≤ μ ≤Interval Width =The tolerance is +/­ 0.005.A random sample of 15 items is taken, producing a sample mean of 2.364 with a sample variance of .81. Assume x is normally distributed and construct a 90% confidence interval for the population mean. Round the answers to 3 decimal places.≤ μ ≤The tolerance is +/­ 0.005.
According to Runzheimer International, the average cost of a domestic trip for business travelers in the financial industry is \$1,250. Suppose another travel industry research company takes a random sample of 51 business travelers in the financial industry and determines that the sample average cost of a domestic trip is \$1,192, with a sample standard deviation of \$279. Construct a 98% confidence interval for the population mean from these sample data. Assume that the data are normally distributed in the population. Now go back and examine the \$1,250 figure published by Runzheimer International. Does it fall into the confidence interval computed from the sample data? Whatdoes it tell you?Round your answers to 2 decimal places, the tolerance is +/­0.01.The 98% confidence interval: ≤ μ ≤The figure given by Runzheimer International falls (outside or inside) the confidence interval. Therefore, there is noreason to reject the Runzheimer figure as different from what we are getting based on this sample.According to a survey by Runzheimer International, the average cost of a fast­food meal (quarter­poundcheeseburger, large fries, medium soft drink, excluding taxes) in Seattle is \$4.82. Suppose this figure was based on a sample of 27 different establishments and the standard deviation was \$0.37. Construct a 95% confidence interval for the population mean cost for all fast­food meals in Seattle. Assume the costs of a fast­food meal in Seattle are normally distributed. Using the interval as a guide, is it likely that the population mean is really \$4.50? Why or why not?Round the answers to 4 decimal places.≤ μ ≤Since 4.50 in the interval, we are 95% confident that μ 4.50.The tolerance is +/­ 0.0005.Assuming x is normally distributed, use the following information to compute a 90% confidence interval to estimate .313321
320329
319317
340311
325307
310318
Round the answers to 2 decimal places.≤ μ ≤The tolerance is +/­ 0.05.The marketing director of a large department store wants to estimate the average number of customers who enter the store every five minutes. She randomly selects five­minute intervals and counts the number of arrivals at the store.She obtains the figures 54, 32, 41, 44, 56, 80, 49, 29, 32, and 74. The analyst assumes the number of arrivals is normally distributed. Using these data, the analyst computes a 95% confidence interval to estimate the mean value for all five­minute intervals. What interval values does she get?Round the intermediate values to 2 decimal places. Round your answers to 2 decimal places, the tolerance is +/­0.01.≤ μ ≤A meat­processing company in the Midwest produces and markets a package of eight small sausage sandwiches. The product is nationally distributed, and the company is interested in knowing the average retail price charged for this item in stores across the country. The company cannot justify a national census to generate this information. Based on the company information system’s list of all retailers who carry the product, a researcher for the company contacts 36 of these retailers and ascertains the selling prices for the product. Use the following price data and a population standard deviation of 0.113 to determine a point estimate for the national retail price of the product.Construct a 90% confidence interval to estimate this price.
\$2.232.162.122.011.992.23
\$2.112.312.072.241.872.10
\$2.121.982.172.182.092.08
\$2.202.172.302.182.222.05
\$2.172.142.292.322.152.16
\$2.101.822.192.022.192.26
Round the answers to 3 decimal places.Point Estimate≤ μ ≤The tolerance is +/­ 0.005.A bank officer wants to determine the amount of the average total monthly deposits per customer at the bank. He believes an estimate of this average amount using a confidence interval is sufficient. How large a sample should he take to be within \$200 the actual average with 99% confidence? He assumes the standard deviation of total monthly deposits for all customers is about \$1,000.Round your answer up to the nearest integer.Sample .The tolerance is +/­ 1.Suppose you have been following a particular airline stock for many years. You are interested in determining the average daily price of this stock in a 10­year period and you have access to the stock reports for these years. However, you do not want to average all the daily prices over 10 years because there are several thousand data points, so you decide to take a random sample of the daily prices and estimate the average. You want to be 90% confident of your results, you want the estimate to be within \$3.00 of the true average, and you believe the standard deviation of the price of this stock is about \$12.75 over this period of time. How large a sample should you take?Sample size =A national beauty salon chain wants to estimate the number of times per year a woman has her hair done at a beauty salon if she uses one at least once a year. The chain’s researcher estimates that, of those women who use a beauty salon at least once a year, the standard deviation of number of times of usage is approximately 6. The national chain wants the estimate to be within one time of the actual mean value. How large a sample should the researcher take to obtain a 98% confidence level?Round your answer up to the nearest integer.Sample .The tolerance is +/­ 1.

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