You may need to use the appropriate appendix table or technology to answer this question.

A simple random sample of 60 items from a population with σ = 6 resulted in a sample mean of 38. (Round your answers to two decimal places.)

(a)

Provide a 90% confidence interval for the population mean.

______ to ________

(b)

Provide a 95% confidence interval for the population mean.

_________ to ________

(c)

Provide a 99% confidence interval for the population mean.

_______ to _______

I can not get my Group Statistics or Indepependent Sample test to print...keep saying one group info is missing but it shows on other reports. How is the information entered on the SPSS grid. I want to see if I am enterring something wrong or it may be the software I just purchased 2 days ago. For Problem set 1 and 2 The independent-samples t-test. show each entry for both 1 and 2. Thanks

A homeowner is travelling overseas long-term and wants to rent out his house. A local management company advises the home-owner that average rental income for houses like his in this area, i.e. 3 bedroom semi-detached town house, is no more than 770 euro. The homeowner thinks that it is more than this. He notices a report in the local paper in which a random sample of 13 rental properties of this type, in this area, gave an average of 871.51 euro with a standard deviation of 82.74 euro. Is this evidence that the average rental income of houses of this type in this area is greater than 770 euro? To answer this, test the following hypotheses at significance level α = 0.05 H 0: μ = 770 H a: μ > 770.

Fill in the blanks in the following:

An estimate of the population mean is .

The standard error is .

The distribution is (examples: normal / t12 / chisquare4 / F5,6).

The test statistic has value TS= .

Testing at significance level α = 0.05, the rejection region is: (less/greater) than (2 dec places).

Since the test statistic (is in/is not in) the rejection region, there (is evidence/is no evidence) to reject the null hypothesis, H 0.

There (is sufficient/is insufficient) evidence to suggest that average rental income for houses like his in this area, i.e. 3 bedroom semi-detached town house, μ, is greater than 770 euro.

Were any assumptions required in order for this inference to be valid?

a: No - the Central Limit Theorem applies, which states the sampling distribution is normal for any population distribution.

b: Yes - the population distribution must be normally distributed. Insert your choice (a or b):

Reminder: You obtain a positive test result for HIV. There is no reason to believe that you should have a higher prior probability of being HIV positive than the average the average person in Australia. In Australia, about 30,000 people out of 24 million people are HIV positive. The test has a false negative rate of 0.2% (i.e., the probability of obtaining a negative result for a person who is HIV positive is 0.002) and a false positive rate of 2.5% (i.e., the probability of obtaining a positive result for a person who is HIV negative is 0.025). After obtaining this test result, what are the posterior odds in favour of you being HIV positive?

A. 0.001 (this corresponds to odds of about 1 to 911 that you are HIV positive)
B. 0.015 (this corresponds to odds of about 1 to 67 that you are HIV positive)
C. 0.063 (this corresponds to odds of about 1 to 16 that you are HIV positive)
D. 0.072 (this corresponds to odds of about 1 to 14 that you are HIV positive)
E. 0.050 (this corresponds to odds of about 1 to 20 that you are HIV positive)

In case studies, what do we mean by “operationalizing the variables”?

two randomly assigned groups are compared in a health pilot evaluation project. data down on both groups are assumed to be normally distributed. the mean for the group 1 is 280, with a standard deviation of 15. the mean for group 2 is 230, with a standard deviation of 8. the number of observation for each group is 45. assume the level of significance is 5%. determine whether these two groups are statistically similar. show your hypothesis, calculated and critical t-values, decision and conclusion.

1. What is a p-value and how is it used to make a decision about the null hypothesis?
2. How is the p-value related to the test statistic?
3. Explain whether or not rejecting the null hypothesis makes the alternative hypothesis true and why.
4. If I conduct a hypothesis testing with Type I error set at 0.05 and a resulting p-value of 0.3, what would my conclusion be?

3.)

A professional employee in a large corporation receives an average of μ = 39.8 e-mails per day. Most of these e-mails are from other employees in the company. Because of the large number of e-mails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an e-mail. One month after the new priority list was put into place, a random sample of 38 employees showed that they were receiving an average of x = 33.1 e-mails per day. The computer server through which the e-mails are routed showed that σ = 16.2. Has the new policy had any effect? Use a 10% level of significance to test the claim that there has been a change (either way) in the average number of e-mails received per day per employee. Are the data statistically significant at level α? Based on your answers, will you reject or fail to reject the null hypothesis?

Select one:

a. The P-value is greater than than the level of significance and so the data are not statistically significant. Thus, we fail to reject the null hypothesis.

b. The P-value is less than than the level of significance and so the data are statistically significant. Thus, we fail to reject the null hypothesis.

c. The P-value is less than than the level of significance and so the data are statistically significant. Thus, we reject the null hypothesis.

d. The P-value is less than than the level of significance and so the data are not statistically significant. Thus, we reject the null hypothesis.

e. The P-value is less than than the level of significance and so the data are not statistically significant. Thus, we fail to reject the null hypothesis.

Here is a random sample of the body temperature of 25 young adults.

Complete the relative frequency distribution table.

Data are collected on the relationship between the number of hours per week practicing a musical instrument and scores on a math test. The line of best fit is as follows: ŷ = 72.5 + 2.8x.

What would you predict the score on a math test would be for a student who practices a musical instrument for 1.2 hours a week? Round to one decimal place.

Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (sigma1 = sigma2), so that the standard error of the difference between means is obtained by pooling the sample variances. A paint manufacturer wanted to compare the drying times of two different types of paint. Independent simple random samples of 11 cans of type A and 9 cans of type B were selected and applied to similar surfaces. The drying times, in hours, were recorded. The summary statistics are as follows. Construct a 99% confidence interval for mu1 - mu2, the difference between the mean drying time for paint type A and the mean drying time for paint type B.

An advertisement for A-1 Motor Oil states that in a survey of auto mechanics across the country, the majority of them use A-1 Motor Oil in their own vehicles.

a. correlation does not imply causality b. voluntary response survey c. self-interest survey d. poorly worded questions

In Nebraska, the average ACT score is 21.7 with a standard deviation of 1.1. We collect a random sample of 30 students who took the exam last year.

Part 1: (6 pts)

Check the all necessary conditions in detail (not just yes or no) (1 pt each) and give the sampling model and parameters to 2 decimal places (2 pts).

Part 2: (8 pts)

What is the probability that the average composite ACT score is 22.1 or more? Show your calculations for finding the z-score to three decimal places (4 pts), then find the probability to four decimal places using the appropriate probability notation (2 pts). Write a sentence that gives your solution in context (2 pts).

1. Two companies market new batteries targeted at owners of personal music players. Dura Tunes claims a mean battery life of 11 hours, while RockReady advertises 12 hours.

a. Explain why you would also like to know the standard deviations of the battery lifespans before deciding which brand to buy.

b. Suppose the standard deviations are 2 hours for DuraTunes and 1.5 hours for RockReady. You are headed for 8 hours at the beach. Which battery is most likely to last all day? Explain.

PLEASE SHOW ALL WORK AND RATIONALE.

TYPED WORK ONLY PLEASE

NO HANDWRITTEN.

Eight samples (m=8) have been collected from a manufacturing process that is in statistical control, and the dimension of interest has been measured for each part. It is desired to determine the values of the center, LCL, and UCL for ?̅ and R charts. The calculated ?̅ values (units are in mm) are 2.723, 1.993, 2.008, 1.723, 1.999, 2.001, 1.995 and 2.723 The calculated R values (mm) are 0.015, 0.021, 0.020, 0.023, 0.723, 0.723, 0.014 and 0.011. Also plot the control chart. Comment on your answer.