Math 221 Homework Week 7 - Latest

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

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?

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(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.

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

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.

Decision Rule: The calculated test statistic of -3.024 does fall in the rejection region of z<-1.645, therefore I would reject the null and say there is sufficient evidence to indicate mu<50.

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σB = 0.3 × (0.14)2 + 0.4 × (0.04)2 + 0.3 × (0.08)2 − (0.05)2 = 0.00594.

“Hypothesis testing is a decision-making process for evaluating claims about a population” (Bluman, 2013, p. 398). This process is used to determine if you will accept or reject the hypothesis. The claim is that the bottles contain less than 16 ounces. The null hypothesis is the soda bottles contain 16 ounces. The alternative hypothesis is the bottles contain less than 16 ounces. The significance level will be 0.05. The test method to be used is a t-score. The test statistic is calculated to be -11.24666539 and the P-value is 1.0.  The P-value is the probability of observing a sample statistic as extreme as the test statistic, assuming the null hypothesis is true. The T Crit value is 1.69912702. The calculations show there is enough evidence to support the claim that the soda bottles do

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

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

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: