Use the data below to test the claim that the weight loss was greater with the low fat diet compared to the low carbohydrate diet. Usea 5% level of significance.Sample DataLow FatLow CarbohydrateT1 = 4.7lbs1 = 7.2lbn1 = 77T1 = 2.6lb81 = 5.9lbnị = 791. Identify the tail of the test.[ Select ]2. What is the P-value? [ Select ]3. Will the null hypothesis be rejected? [Select ]4. Is the initial claim supported? [ Select ]

The following information has been provided Sample mean 1=4.7 Sample standard deviation 1=7.2…

Answered: Use the data below to test the claim…

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 data below to test the claim that the weight loss was greater with the low fat diet compared to the low carbohydrate diet. Use a 10% level of significance. Low Fat Low Carbohydrate x¯1=3.7lbs 1=5.2lb n1=57 x¯1=2.7 lbs1=4.9lb n1=62 Identify the tail of the test. What is the P-value? Will the null hypothesis be rejected? Is the initial claim supported?
Assume that you want to test the claim that the paired sample data below come from a population from which the mean difference is zero. Using JMP, compute the value of the appropriate test statistic. Do the data appear to come from a population where the mean difference is zero? X: 28 35 25 25 32 30 28 37 Y: 26 31 31 25 33 35 28 33 a. No b. O c. Yes d.
An operations manager wants to know whether the maintenance of a filling machine for 2 litre milk bottles has made a difference in the average fill of the milk bottle. She randomly selects 25 bottles before the maintenance service, and 25 bottles after the maintenance service has taken place. The resulting SPSS output is shown below: Paired Samples Statistics Mean N Std. Deviation Std. Error Mean Pair 1 After Service 2.0029 25 .01142 .00228 Before service 1.9687 25 .03193 .00639 Paired Samples Test Paired Differences t df Sig. (2- tailed) Mean Std. Deviation Std. Error Mean 95% Confidence Interval of the Difference Lower Upper Pair 1 After Service – Before service .03420 .03503 .00701 .01974 .04866 4.882 24 .000 Question: Based on the output choose the most accurate answer
After conducting a re-test for independent samples, you arrived at the following:t(47) = 3.41, p<.05. The degrees of freedom is indicated by which of the following? A. 47 B. 3.41 C. <.05 D. P E. T
What is the sample size for the given data
Listed in the accompanying table are heights (in.) of mothers and their first daughters. The data pairs are from a journal kept by Francis Galton. Use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal. Use a 0.05 significance level to test the claim that there is no difference in heights between mothers and their first daughters. Mother 64.0 66.0 63.0 62.0 66.5 66.0 65.0 60.0 67.0 63.0 Daughter 67.0 66.5 70.5 66.0 61.0 66.0 65.5 65.0 67.0 65.0 In this example, Hd is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the daughter's height minus the mother's height. What are the null and alternative hypotheses for the hypothesis test? Ho: Ha = 0 in. H₁ Hd 0 in. (Type integers or decimals. Do not round.) Identify the test statistic. t= (Round to two decimal places as needed.)
Assume that you want to test the claim that the paired sample data from a for which the mean difference Ha = 0. Give the value of the ttest statistic. 11 13 5 Y 8 6 4 Note: Use three decimal places.
instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores: Grades 63.9 68.4 | 69 | 96 64.3 | 63.9 65 69 62.7 68.4 Test grades are believed to be normally distributed. Use a significance level of 5%. A. State the alternative hypothesis: HA: Ou + 65 Ομ 65 B. State the mean of the sample: C. State the standard error of the sample means:
The accompanying data are the weights (kg) of poplar trees that were obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the accompanying table. Also shown are partial results from using the Bonferroni test with the sample data. No treatment Fertilizer Irrigation Fertilizer and Irrigation 1.26 1.02 0.09 0.81 0.47 .86 .52 1.71 0.45 0.55 0.06 1.45 0.14 0.67 0.79 2.37 1.33 0.97 0.76 1.66 (I) TREATMENT (J) TREATMENT Difference (I-J) Std. Error Sig. 1.00 2.00 −0.0840 0.28375 1.000 3.00 0.2860 0.28375 1.000 4.00 −0.8720 0.28375 0.037 1. Use a 0.10 significance level to test the claim that the different treatments result in the same mean weight. a. Determine the null and alternative hypotheses. Find the test statistic. ______ (Round to two decimal places as needed.) Find the…
Listed in the accompanying table are heights (in.) of mothers and their first daughters. The data pairs are from a journal kept by Francis Galton. Use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal. Use a 0.05 significance level to test the claim that there is no difference in heights between mothers and their first daughters. ... Question content area top right Part 1 Mother 62.0 65.0 64.7 65.5 65.0 67.0 66.0 66.5 63.0 58.5 Daughter 68.0 69.0 66.5 63.0 68.0 62.0 66.5 66.7 63.5 66.5 Question content area bottom Part 1 In this example, μd is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the daughter's height minus the mother's height. What are the null and alternative hypotheses for the hypothesis test? H0:…
Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is Hd = 0. Compute the value of the t test statistic. Round intermediate calculations to four decimal places as needed and final answers to three decimal places as needed. 28 31 20 25 28 27 33 35 y 26 27 26 25 29 32 33 34 X OA. t= -1.185 OB. t= -0.523 OC. t= -1.480 OD. t= -0.690 2 PE 7 2 W Z S x H mmand I E D с 4 1 R F D % L T G 6 B Y 3 I H N U J . 8 I M ( 9 K I O H 2 O L P command Time Remaining: 01:12:18 ; x ( option ? "1 I . Next return
How much customers buy is a direct result of how much time they spend in the store. A study of average shopping times in a large national houseware store gave the following information (Source: Why We Buy: The Science of Shopping by P. Underhill). Women with female companion: 8.3 min.Women with male companion: 4.5 min. Suppose you want to set up a statistical test to challenge the claim that a woman with a female friend spends an average of 8.3 minutes shopping in such a store. (a) What would you use for the null and alternate hypotheses if you believe the average shopping time is less than 8.3 minutes? Ho: ? = 8.3; H1: ? ≠ 8.3Ho: ? < 8.3; H1: ? = 8.3 Ho: ? = 8.3; H1: ? < 8.3Ho: ? = 8.3; H1: ? > 8.3 Is this a right-tailed, left-tailed, or two-tailed test? right-tailedtwo-tailed left-tailed (b) What would you use for the null and alternate hypotheses if you believe the average shopping time is different from 8.3 minutes? Ho: ? = 8.3; H1: ? > 8.3Ho: ? ≠ 8.3; H1: ? = 8.3…

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