Hypothesis Tests of a Single Population

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?

(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.

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.

Is this estimate centered about the parameter of interest (the parameter of interest is the answer for the mean in question 2)?

12. _____ For a given population, confidence intervals constructed from larger samples tend to be narrower than those constructed from smaller samples. Which statement below best describes why this is true? (A) The variability of the sample mean is less for larger samples. (B) The z-value for larger samples tends to be more accurate. (C) The population variance is larger for large populations. (D) As the sample size increases, the z-value (or t-value) becomes smaller. A machine dispenses potato chips into bags that are advertised as containing one pound of product. To be on the safe side, the machine is supposed to be calibrated to dispense 16.07 ounces per bag, and from long time observation, the distribution of the fill-weights is known to be approximately normal and the process is known to have a standard deviation of 0.15 ounces.

H_a: μ 30 we will use a z-test for the mean to test the given hypothesis.

Because it is one sided, α will remain .05. Thus, comparing .0384 to .05, it is lower. In conclusion, the mean age of stars is indeed greater than 3.3. We reject the null hypothesis.

46. As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 220 applicants 14 failed the test. Develop a 99 percent confidence interval for the proportion of applicants that fail the test. Would it be reasonable to conclude that more than 10 percent of the applicants are now failing the test? In addition to the testing of applicants, Fashion Industries randomly tests its employees throughout the year. Last year in the 400 random tests conducted, 14 employees failed the test. Would it be reasonable to conclude that less than 5 percent of the employees are not able to pass the random drug test?

1. A researcher is interested in whether students who attend private high schools have higher average SAT Scores than students in the general population. A random sample of 90 students at a private high school is tested and and a mean SAT score of 1030 is obtained. The average score for public high school student is 1000 (σ= 200).

Imagine you are a manager at a major bottling company. Customers have begun to complain that the bottles of the brand of soda produced in your company contain less than the advertised sixteen (16) ounces of product. Your boss wants to solve the problem at hand and has asked you to investigate. You have your employees pull thirty (30) bottles off the line at random from all the shifts at the bottling plant. You ask your employees to measure the amount of soda there is in each bottle.

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

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.

Do Major League Baseball teams with higher salaries win more frequently than other teams? Although many people believe that the larger payroll budgets win games, which point does vary, depending on the situation. “…performances by individual players vary quite a bit from year to year, preventing owners from guaranteeing success on the field. Team spending is certainly a component in winning, but no team can buy a championship.” (Bradbury). For some, it’s hard not to root for the lower paid teams. If the big money teams, like Goliath, are always supposed to win, it’s hard not cheer for David. This paper will discuss the effects of payroll budgets on the percentage of wins for the 30 Major League