Math 221 Homework Week 7 - Latest

σA = 0.3 × (0.07)2 + 0.4 × (0.06)2 + 0.3 × (0.08)2 − (0.021)2 = 0.004389,

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Let’s assume you have taken 1000 samples of size 64 each from a normally distributed population. Calculate the standard deviation of the sample means if the population’s variance is 49.

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11. The frequency histogram below describes the distribution of the lifetimes of 220 light bulbs.

(21) You took a sample of size 21 from a normal distribution with a known standard deviation, . In order to find a 90% confidence interval for the mean, You need to find.

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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

In this project we were given the case of customer complaints that the bottles of the brand of soda produced in our company contained less than the advertised sixteen ounces of product. Our boss wants us to solve the problem at hand and has asked me to investigate. I have asked my employees to pull Thirty (30) bottles off the line at random from all the shifts at the bottling plant. The first step in solving this problem is to calculate the mean (x bar), the median (mu), and the standard deviation (s) of the sample. All of those calculations were easily computed in excel. The mean was computed by entering:

auto insurance offered by two leading companies. He selects a sample of 15 families, some with only a single insured driver, others with several teenage drivers, and pays each family a stipend to contact the two companies and ask for a price quote. To make the data comparable, certain features, such as the deductible amount and limits of liability, are standardized. The sample information is reported below. At the .10 significance level, can we conclude that there is a difference in the amounts quoted?

2) Compute the standard deviation for each of the four samples. Does the assumption of .21 for the population standard deviation appear reasonable?

21) Seventy two percent of all observations fall within 1 standard deviation of the mean if the data is normally distributed.

Conclusion : Reject the null hypothesis. The sample provide enough evidence to support the claim that mean is significantly different from 12 .