Week 7 Quiz Notes

The area under the curve to the left of the unknown quantity must be 0.7 (70%). So, we must first find the z value that cuts off an area of 0.7 in the left tail of standard normal distribution. Using the cumulative probability table, we see that z=0.53.

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?

(c) The mean of the sample and the value of Z with an area of 5% in the left tail.

9. Assume the weight, W, of a randomly selected adult has a normal distribution with a mean

The mean for the column “mean” is 3.56. It is very close to the parameter of interest but is not equal to it. You can calculate a confidence interval for the mean of the mean column, but a specific confidence interval would need to be provided. In that case, the confidence interval would be centered on 3.56, not 3.5.

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.

Complete the calculations below using this data. Show all of your work and clearly label each of your calculations.

standard deviation standardized value rescaling z-score normal model parameter statistic standard Normal model 68-95-99.7 Rule normal probability plot

2. For the following set of scores, fill in the cells. The mean is 74.13 and the standard deviation is 9.98.

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.

Fry Brothers heating and Air Conditioning, Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. A random sample of 40 days last year showed that Larry Clark made an average of

The 95% confidence interval for the population mean is 66,438 to 80,241. This means that there is a 95% confidence that this interval has the population mean.

– (The above are two ways to write the same equation, the first allows you to calculate z from Q and the second lets you calculate Q from z.) – Look up Prob{the outcome of a standard normal is z or lower} in the Standard Normal Distribution Function Table. utdallas.edu/~metin

(Points: 5.0)     Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis.

Z = the standard normal deviate set at 1.96 which corresponds to the 95 % confidence level.