A data set includes the counts of chocolate chips from three different types of Chips Ahoy cookies. The accompanying StatCrunch display shows results fromanalysis of variance used with those three types of cookies. Use a 0.05 significance level to test the claim that the three different types of cookies have thesame mean number of chocolate chips.ANOVA tableSourceColumnsErrorTotalDetermine the null hypothesis.Ho:DFSSMSF-Stat P-value2 1223.9500 611.97500 62.3968 <0.000177 755.20000 9.807792279 1979.1500Determine the alternative hypothesis.H₁:Determine the test statistic.The test statistic is(Round to two decimal places as needed.)Determine the P-value.The P-value is(Round to three decimal places as needed.)What can you conclude?Theresufficient evidence at a 0.05 significance level to warrant rejection of the claim that the three different types of chocolate chip cookies havemean number of chocolate chips.

Given that dftreatment = 2 dferror = 77 dftotal = 79

Answered: A data set includes the counts of…

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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4. Answer the following. Feel free to use the Excel Demo from the course website called "alpha and beta". a. What happens to the probability of type IlI error as the difference between mu_0 and the actual mean increases? b. What happens to the probability of type II error as the sample size increases? C. How does changing the significance level effect the type Il error probability? -Accessibility: Good to go O Focus a AP hp
Do campers with trailers really get better mileage per gallon on the highway? The table shows results from a study of the MPG (miles per gallon) of campers with trailers both in the mountains and on desert roads. Assume that the two samples are randomly selected, independent, the population standard deviations are not know and not considered equal. At the 0.05 significance level, test the claim that the mpg on desert roads is better than in the mountains. MPG on Desert Roads 30.9 32.6 28.4 25.4 35.1 32.4 31.1 30.9 34.1 35.5 MPG in the Mountains 24 21.2 22.3 24.9 23.2 24.2 25.1 25.8 23.1 28.1 What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) Hg: Select an answer v Select an answer H: Select an answer ? v Select an answer v Original Claim = Select an answer v df =
9. Summarize the results from an analysis (two-way ANOVA) performed on the data graphed below. Assume big differences are significant and trivial differences are not. Mean y 22.5- 20.0- 17.5 15.0 12.5 10.0
Big babies: The National Health Statistics Reports described a study in which a sample of 96 one-year-old baby boys were weighed. Their mean weight was 25.6 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys is greater than 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the =α0.05 level of significance and the P -value method and Excel. Part: 0 / 5 0 of 5 Parts Complete Part 1 of 5 (a)State the appropriate null and alternate hypotheses. H0 : =μ25 H1 : >μ25 This hypothesis test is a ▼two-tailed test. Part: 1 / 5 1 of 5 Parts Complete Part 2 of 5 (b)Compute the value of the test statistic. Round the answer to at least three decimal places.…
Medicare spending per patient in different U.S. metropolitan areas may differ. Based on the sample data below, answer the questions that follow to determine whether the average spending in the northern region is significantly less than the average spending in the southern region at the 1 percent level. 1) Specify decision rule. Reject the null hypothesis if tcalc <____ 2) Find the test statistic tcalc assuming unequal variances Medicare Spending per Patient (adjusted for age, sex, and race) Statistic Northern Region Southern Region Sample mean $4324 $7,426 Sample standard deviation $1,537 $3,495 Sample size 11 patients 18 patients
Report an appropriate hypothesis test and use a 5% significance level. The alternative hypothesis for the One-way Anova F-test would be Group of answer choices Ha: Density affects firing temperature. Ha: Firing temperature affects density. Ha: Firing temperature does not affect density. Ha: Density does not affect firing temperature.
Compute a one-way ANOVA on the following data using 0.05 significance level. Group A Group B Group C 14 11 14 13 12 12 10 13 13 12 13 12 14 ANOVA Source of Variation SS df MS F P-value F crit Between Groups ? ? ? ? ? Within Groups ? ? ? Total ? ?
It is known that the amount of glass households in Phoenix recycled 10 years ago is normally distributed with a mean of 3.6 pounds of glass per week. Phoenix municipal waste recently chose a simple random survey of 24 households and found they recycled an average of 3.8 pounds a week with a standard deviation of 0.5 pounds. Using a significance level of 0.05, is this evidence that the amount of glass recycling has changed? Write a one or two sentence summary in non-statistical language to describe what you have found, not how you found it. Your audience is people who have never had a statistics class (think of middle of the road 8th graders, not the really bright ones). Is there anything about the sample data suggesting that the methods we learned should not be applied?
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A bottle contasins a tabel stating that it contains pills with 500 mg of vitamin C, and another bottle contains a label stating that it contains pills with 325 mg of aspirin. When testing claims about the mean contents of the pills, which wouid have mo serious implications rejection of the vitamin C olaim or rejection of the aspirin claim? Considering only a type l error and using the same sample size, is it wise to use the same significance level for hypothenis tests about the maan amount of vitamin Cand the mean amount of aspirin? Rejection of the clam about claim about the anpinn. V more senious because the wrong dosage could cause more serious adverse reanctions than a wrong dosage. It would be wise to use a V significance level for testing the
A cold medicine lists 600 milligrams of acetaminophen per fluid ounce as an active ingredient. A test of 75 one-ounce samples of the medicine finds that the mean amount of acetaminophen for the sample is 598 milligrams with a standard deviation of 20 milligrams. Test the claim that the medicine does not contain the required amount of acetaminophen. Use a 0.05 significance level. State the null and alternative hypotheses. O A. Ho: μ = 600 milligrams H₂:μ>600 milligrams OC. Ho: μ = 600 milligrams H₂: μ< 600 milligrams Find the z-score. Z= (Round to two decimal places as needed.) B. Ho: μ = 600 milligrams H₂: μ#600 milligrams O D. Ho: μ#600 milligrams H₂: μ = 600 milligrams M-I

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