Quiz: Standard Deviation and Confidence Interval Estimate

2. In order to determine the average amount spent in November on Amazon.com a random sample of 144 Amazon accounts were selected. The sample mean amount spent in November was $250 with a standard deviation of $25. Assuming that the population standard deviation is unknown, what is a 95% confidence interval for the population mean amount spent on Amazon.com in November?

(2) Give that a sample of 25 had x = 75, and (x-x)² = 48 the mean and standard

10 . Eighteen percent of men prefer brand A of a certain product. Suppose 5 men are

A business wants to estimate the true mean annual income of its customers.  It randomly samples 220 of its customers.  The mean annual income was $61,400 with a standard deviation of $2,200.  Find a 95% confidence interval for the true mean annual income of the business’ customers.

Body weight expressed as a percentage of ideal in an SRS of n = 9 girls selected at random are as follows {114, 100, 104, 94, 114, 105, 103, 105, 96}. Calculate a 95% confidence interval for the population mean µ.

g. Construct a 97% confidence interval estimate of the population proportion of all organic tea bags produced by Ryerson Inc., which have more than 5.00 gram.

The customers in this case study have complained that the bottling company provides less than the advertised sixteen ounces of product. They need to determine if there is enough evidence to conclude the soda bottles do not contain sixteen ounces. The sample size of sodas is 30 and has a mean of 14.9. The standard deviation is found to be 0.55. With these calculations and a confidence level of 95%, the confidence interval would be 0.2. There is a 95% certainty that the true population mean falls within the range of 14.7 to 15.1.

Calculate the 95% confidence interval for the population mean weekday sleep time of all adult residents of this Midwestern town.

A pharmaceutical company is testing the effectiveness of a new drug for lowering cholesterol. As part of this trial, they wish to determine whether there is a difference between the effectiveness for women and for men. Using   = .05, what is the value the test statistic?

I chose (c) for the answer as I thought that since both schools took the same percentage size of about 3% of the undergraduates and both had a proportion, p= 0.80, Johns Hopkins and Ohio State would have almost the same sampling variability. I forgot to take into account that each school had different numbers of undergraduates. However, the correct answer is (b). Since a simple random sample of about 3% of Ohio State is about (0.03 x 40,00) = 1,200, it has a larger sample than the (0.03 x 2,000) = 60 undergraduates that make up the simple random sample of about 3% of the undergraduates at Johns Hopkin, meaning Ohio State has smaller variability. Thus, the estimate from Johns Hopkins has more sampling variability than that from Ohio State since it has a smaller sample of 60 undergraduates, making

d) Among all employees whose starting salary is below the median ($37,750), find a 95% confidence interval for the proportion who stay with D&Y for at least 3 years.

How many repair records should be sampled if the research firm wants the population mean number of miles driven until transmission failure to be estimated with the margin of error of 5000 miles? Use 95% confidence.

b. However, none of us were born in 1932 so that percentage decreases to less than 2.6 percent for those born in 1976 (Heritage Foundation, 2000).

a. Conclusion drawn must be based on a sample that represent the entire group .

orange juice to complete a survey on a new line of frozen orange juice. After