T-tests were done on the means to determine of the differences between the means.t-statt-crit1.027762.3646242.4469122.364624FF1 VS FF2FF1 VS FF33.5527035.360156FF2 VS FF3Statement 1: The mean calories in the sandwiches from Fast Food 1 is the same as the mean calories in thesandwiches from Fast Food 2.Statement 2: The mean calories in the sandwiches from Fast Food 1 is the same as the mean calories in thesandwiches from Fast Food 3.Statement 3: The mean calories in the sandwiches from Fast Food 2 is the same as the mean calories in thesandwiches from Fast Food 3.O All statements are FALSE.O Only Statement 2 is true.O Only Statements 2 & 3 are true.O All statements are TRUE.O Only Statement 3 is true.O Only Statements 1 & 2 are true.O Only Statements 1 & 3 are true.O Only Statement 1 is true.

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Answered: T-tests were done on the means to…

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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Use the 1odwing Wildlife: Coyotes A random sample of 46 adult coyotes in a region of northern Minnesota showed the average age to be z = 2.05 years, with sample standard deviation 8 = 0.82 years (based on information from the book Coyotes: Biology, Behavior and Management by M. Bekoff, Academic Press). However, it is thought that the overall population mean age of coyotes is µ=1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use a = 0.01. Fl in the blanks a 007 • Ho u- 175 years * H u> 176 yean QUESTION 7 What is your test statistic? (t-statistic) 2.48 QUESTION 8 What is your p-value? (Hint You will need the degrees of freedom to find the p-value.) <0.01d QUESTION 9 Do you reject the null hypothesis? O Yes No
Á study was conducted that examined the relationship between alcohol consumption and risk of heart attack. In all, 417 men, age 40 or older, were tracked over ten years and the number of that experienced a heart attack was recorded. The participants were grouped by those who totally abstained from alcohol and those that drank moderately. For the results given in the table below, use the chi-squared test to find the p value for NHST at the .05 level. (In SPSS use the Yates continuity correction value.) Heart attack Alcohol Crosstabulation Count Alcohol Abstainer Moderate Total Heart attack No 197 192 389 Yes 9. 19 28 Total 206 211 417 Op= .090 Op= 121 !! Op=.067 Op = .048
The manager of a political party wants to find out the association between the campaign spending (Large, medium and small) and the vote obtained (Majority, Average and minority). Use Chi-squared hypothesis test for association to test that there is an association between the variable in campaign spending and the variable in vote obtained with five percent level of significance. The manager uses MS excel to calculate the Chi-squared stat value and the result is 8.874. Critical value: At 1% = 13.277 :At 5% = 9.488 :At 10% = 7.779 1. State the decision rule 2. Conclusion and decision 3. State the alternative hypothesis and the null hypothesis
Help me out. Thanksss a lot
Is narcissism a more common personality trait today than it was a few decades ago? It is known that the mean population score on the Narcissistic Personality Inventory (NPI) for students attending University of South Alabama around 20 years ago was μ= 15 (Twenge, 2010). Interested in the narcissism levels of students in the year 2020, a researcher administers the NPI to a random sample of 25 University of Alabama sophomores this Spring term. The mean NPI score from the researcher’s sample of sophomores is M = 16.5, with s = 3.4. 1. Write the null and alternative hypotheses in symbols. Possible symbols for your answer: H0, H1, μ, M, σ. 2. Calculate the standard error. 3. Find the critical value for the test statistic, assuming alpha = .05 (Use largest [i.e., most conservative] value if exact value not given in the chart) a) 2.064 b) 1.96 c) 1.98 d) 2.000
The proprietor of a greenhouse claims that its pots of lilies average 6.0 buds, with s = 0.4 buds. If n = 29 pots of lilies from a large no of pots of lilies has a mean of 5.8 buds, does this deny the proprietor’s claim of the mean number of buds at 0.015 level of significance?
Suppose the mean retail price oer gallon of regular gasoline int the United States is $3.42 with a standard eviation of $0.40 and the retail price per gallon has a normal bell shaped distribution. Use the Empirical Rule to answer the following quesitona. What percentage of regular gasoline sold between $2.62 and $3.82 per gallon? what percentage of regualr gasoline sold for more the $3.82 per gallon?
One important consideration in determining which location is best for a new retail business is the amount of traffic that passes the location each business day. Counters are placed at each of four locations on the five weekdays, and the number of cars passing each location is recorded in the table below. At 0.05 level of significance, is there sufficient evidence of a significant difference in the mean number of cars per day at the four locations?Using MS Excel,what is the computed value of the F statistic ? * Location II II IV 45 48 44 39 50 60 50 49 39 40 38 39 44 45 42 40 42 43 44 43 0.76 2.23 1.19 0.84
Data from 500 random tornados was collected and the length in miles that the tornado was touching the ground was recorded. We want to test the claim that the mean tornado length is greater than 2.5 miles using a .05 significance level. Using the traditional method if the test statictic is .759, x hat = 2.72424 and s= .2956. Tell why you made the decision you did.
The average house has 13 paintings on its walls. Is the mean different for houses owned by teachers? The data show the results of a survey of 12 teachers who were asked how many paintings they have in their houses. Assume that the distribution of the population is normal. 12, 14, 14, 14, 15, 14, 14, 14, 12, 13, 15, 14 What can be concluded at the αα = 0.01 level of significance? For this study, we should use The null and alternative hypotheses would be: H0:H0: H1:H1: The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is αα Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest the population mean is not significantly different from 13 at αα = 0.01, so there is sufficient evidence to conclude that the population mean number of paintings that are in teachers' houses is equal to 13. The data…
What is the Sum of Squares within group? What is the Mean Square between group? What is the Mean Square within group?
Show the 4-Step Method in testing significance of mean difference in the following case:A topic of recent clinical interest is the effect of different contraceptive methods on fertility. Suppose we wish to compare how long it takes users of either oral contraceptives (OCs) or diaphragms to become pregnant after stopping contraception. A study group of 20 OC users is formed, and diaphragm users who match each OC user with regard to age (within 5 years), race, parity (number of previous pregnancies), and socioeconomic status (SES) are found. The investigators compute the differences in time to fertility between previous OC and diaphragm users and find that the mean difference đ (OC minus diaphragm) in time to fertility is 4 months with a standard deviation (sd) of 8 months. Can we conclude that previous OC users take a significantly longer time to become pregnant than do previous diaphragm users?

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