A psychologist has developed a new test to measure depression. Using a very large groupof "normal" individuals, the mean score on this test is found to be p=55 and a populationstandard of o=12. The scores form a normal distribution. A researcher would like to usethis test to evaluate a theory predicting that increased depression is a common part of theaging process. To test the theory, a sample of n=36 elderly people is obtained. Thedepression test is administered to each person, and the mean score for the sample is 59.Testing the hypothesis using a one-tailed test at 0.05 level of significance, do these datasupport the theory? You can answer "Yes" or "No". Do the 4-step process to determine theanswer.Answer:

For the given data Perform Z test for one mean

Answered: A psychologist has developed a new test…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A school reports that 89% of its graduates get jobs within one year. You take a random sample of 53 graduates, of whom 43 got a job within one year. Is this enough evidence to show that a significantly different percent of graduates get jobs than advertised by the school? (Use a=0.10)
You are conducting a study to see if the proportion of voters who prefer candidates a is significantly different from 0.6. Thus you are performing a two-tailed test. Your sample data produce the test statistics z=-1.65. Find the p-value accurate to 4 decimal places
A market research firm used a sample of individuals to rate the purchase potential of a particular product before and after the individuals saw a new television commercial about the product. The purchase potential ratings were based on a 0 to 10 scale, with higher values indicating a higher purchase potential. The null hypothesis stated that the mean rating "after" would be less than or equal to the mean rating "before." Rejection of this hypothesis would show that the commercial improved the mean purchase potential rating. Use a = 0.05 and the following data to test the hypothesis and comment on the value of the commercial. Purchase Rating Individual After Before 1 6 5 6 4 3 7 4 4 3 3 8 8 6 State the null and alternative hypotheses. (Use u, = mean rating after - mean rating before.) O Ho: Hs0 H3: H = 0 O Ho: Hd#0 Hai Hq= 0 O Ho: Hg=0 O Ho: Hd> 0 Calculate the value of the test statistic. (Round your answer to three decimal places.) 1.183 Calculate the p-value. (Round your answer to…
A market research firm used a sample of individuals to rate the purchase potential of a particular product before and after the individuals saw a new television commercial about the product. The purchase potential ratings were based on a 0 to 10 scale, with higher values indicating a higher purchase potential. The null hypothesis stated that the mean rating "after" would be less than or equal to the mean rating "before." Rejection of this hypothesis would show that the commercial improved the mean purchase potential rating. Use a = .05 and the following data to test the hypothesis and comment on the value of the commercial. Purchase Rating Purchase Rating Individual After Before Individual After Before 1 6. 6. 3 2 6. 4 6. 7 8 7 6. 5 4 4 8 6. 6 a. What are the hypotheses? Ho: ld is Select Ha: ld is select b. Compute d (to 3 decimals). Compute sa (to 1 decimal).
You recently had a cholesterol panel completed and see that the results for your High Density Lipoprotien (HDL) level comes back with z = -1.7 among people of your stature. Your doctor is going to review the results with you but based on what you know about z-scores you can infer: a.The average person of your stature is 1.7 deviations below you. b.Your HDL level is slightly below average. c.The score is negative so this will be good news. d.Your HDL level is extremely low for a person of your stature.
Citrus Rental is a popular car rental agency that has a history of having too few cars available, so that its available cars are overdriven. The mean monthly mileage over the years for Citrus cars has been about 1550 miles per month. Recently, though, Citrus purchased thousands of new cars, and the company claims that the average mileage of its cars is now less than in the past. To test this, a random sample of 15 recent mileages of Citrus cars was taken. The mean of these 15 mileages was 1517 miles per month, and the standard deviation was 201 miles per month. Assume that the population of recent monthly mileages of Citrus cars is normally distributed. At the 0.05 level of significance, can it be concluded that the mean recent monthly mileage, μ, of Citrus cars is less than 1550 miles per month? Perform a one-tailed test. Then complete the parts below.
Citrus Rental is a popular car rental agency that has a history of having too few cars available, so that its available cars are overdriven. The mean monthly mileage over the years for Citrus cars has been about 1500 miles per month. Recently, though, Citrus purchased thousands of new cars, and the company claims that the average mileage of its cars is now less than in the past. To test this, a random sample of 11 recent mileages of Citrus cars was taken. The mean of these 11 mileages was 1332 miles per month, and the standard deviation was 232 miles per month. Assume that the population of recent monthly mileages of Citrus cars is normally distributed. At the 0.05 level of significance, can it be concluded that the mean recent monthly mileage, u, of Citrus cars is less than 1500 miles per month? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary,…
The Department of Agriculture wants to determine whether the mean yield per acre for a particular variety of soybeans is different this year compared to the historical avg yield of 520 bushels per acre. Thus, the null and alternative hypotheses are Ho: mu = 520 Ha: mu is not = to 520 You set out to conduct a hypothesis test at the 95% level. Mean yields are distributed Normally. You take a sample of 36 different acres. From this sample, you calculate a mean yield of 573 and a std deviation of 24. Do you reject or not-reject the null hypothesis? Group of answer choices Do not reject the null hypothesis. Reject the null hypothesis.
Imagine that a researcher wants to compare the intelligence levels of Poodles to all dogs. He hypothesizes that Poodles will score differently than other dogs. He uses a standard test of dog intelligence in his study. The population of all dogs has a mean of μ = 100, and an SD of 15. He obtains a random sample of 500 Poodles and he finds that the mean intelligence score of the sample is 117. Is this a one- or two- tailed test? Write the Ho and Ha for this research question. What kind of statistic should be calculated to test the hypothesis? Compute that statistic by hand What should the researcher conclude? Write a blurb describing the results
Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the weight increase (in pounds) for expectant mothers in the second trimester. In a fictitious study, suppose that CHDS finds the average weight increase in the second trimester is 14 pounds. Suppose also that, in 2015, a random sample of 40 expectant mothers have mean weight increase of 15.8 pounds in the second trimester, with a standard deviation of 6.2 pounds. A hypothesis test is done to see if there is evidence that weight increase in the second trimester is greater than 14 pounds. Find the p-value for the hypothesis test. The p-value should be rounded to 4 decimal places.
Recent results suggest that children with ADHD tend to watch more TV than children who are not diagnosed with the disorder. To examine this relationship, a researcher obtains a random sample of n = 36 children, 8 to 12 years old, who have been diagnosed with ADHD. Each child is asked to keep a journal recording how much time each day is spent watching TV. The average daily time for the sample is M = 4.9 hours. It is known that the average time for the general population of 8 to 12-year-old children is = 4.1 hours with = 1.8. (Hint: Be sure to use the correct test statistic). Are the data sufficient to conclude that children with ADHD watch significantly more TV than children without the disorder? Use a two-tailed test with = .05. (Be sure to give your conclusion, too) If the researcher had used a sample of n = 9 children and obtained the same sample mean, would the results be sufficient to reject H0? (Be sure to give your conclusion, too) Compute Cohen's d for this study. What is…

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