The following statistical test explores gender differences in the numbers of hours spent listening to musiceach day. Rejection of the null hypothesis would imply that:ttest music, by (sex)Two-sample t test with equal variancesGroupMaleFemaleObe33 3.45454554 2.518519472.0735639360269diff mean (Male) sean (Female)combineddiff 1Bo: diff - 0Mean Std. Brr. Std. Dev. (95% conf. Intervall4.6038423.046722Hat diff< 0Pel<) 0.95265642207 3.2412472633451271769255416172.3052491.935183 1.9903152.5348942.333304-165795degrees of freedom-Hat diff 1-0Pri (t) 0.09493.4138222.037849E- 1.609105Hat diff > 0Pr (2> ) - 0.0474O males and females in the population are listening to significantly different amounts of music each day.O the populations of males and females are no different in their standard deviations.O males and females are not listening to different amounts of music each day.O males and females in this sample are listening to significantly different amounts of music each day.

Hypothesis testing is a tool for making statistical inferences about the population data. It is an…

Answered: The following statistical test explores…

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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Given below are the analysis of variance (ANOVA) results from a Minitab display. Assume that you want to use a 0.01 significance level in testing the null hypothesis that the samples come from populations with the same mean. Source DF SS MS F P-val Factor 2. 2280.0 1140.0 7.9 5.39E–04 Error 144 20691.7 143.7 Total 146 22971.8 What can you conclude about the equality of the population means? (circle one) Reject H0 since the p-value is greater than the significance level. Fail to reject H0 since the p-value is greater than the significance level. Reject H0 since the p-value is less than the significance level. Fail to reject H0 since the p-value is less than the significance level.
Listed below are the lead concentrations (in ug/g) measured in different Ayurveda medicines. Ayurveda is a traditional medical system commonly used in India. The lead concentrations listed here are from medicines manufactured in the United States. Assume that a simple random sample has been selected. Use a 0.01 significance level to test the claim that the mean lead concentration for all such medicines is less than 14.0 µg /g. 2.96 6.45 5.99 5.51 20.53 7.45 11.97 20.46 11.52 17.54 D Identify the null and alternative hypotheses. Ho: H1: (Type integers or decimals. Do not round.) Identify the test statistic. (Round to two decimal places as needed.) Identify the P-value. (Round to three decimal places as needed.) State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. V the null hypothesis. There sufficient evidence at the 0.01 significance level to V the claim that the mean lead concentration for all Ayurveda medicines…
The coach of a very popular men’s basketball team claims that the average distance the fans travel to the campus to watch a game is 35 miles. The team members feel otherwise. A sample of 16 fans who travel to games was randomly selected and yielded a mean of M= 36 miles and s= 5 miles. Test the coach’s claim at the 5% (.05) level of significance. one-tailed or two-tailed test: State the hypotheses: df= tα or t value for the critical region = sM = t (test statistic)= Decision:
Ten randomly chosen participants in a Math Training Program took a pretest and a posttest. Test at 0.05 level of significance whether the variance of the posttest scores is significantly different from the variance of the pretest scores. Use the data the following data. [pretest: 5, 6,7,7,7,8,8,9,9,10] [posttest: 6,7,9,8,7,913,10,11,6]
Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. Identify the p-value. Source DF SS MS F Factor 3 30 10.00 1.6 0.264 Error 50 6.25 Total 11 80 O A. 6.25 O B. 1.6 OC. 10.00 O D. 0.264 Click to select your answer. étv MacBook Air F10 F9 F7 30 888 *- F6 F3 esc F1 F2 & @ 7 4 5 2 3.
A researcher wants to test whether the mean height for men and women are different, using a significance level of 0.05. Which is the correct hypothesis? A Ho: P1 = P2 Ha: P1 P2 D Ho: P1 = μ2 Ha: H1 H2 O C A OF OE O B B Ho: P1 = P2 Ha: P1 P2 E Ho: P1 = μ2 Ha: 12 C Ho: P1 = P2 Ha: 1: P₁ P2 F H₂:₁ = 1₂ Ha: ₁ 2
Listed below are the lead concentrations (in µg/g) measured in different Ayurveda medicines. Ayurveda is a traditional medical system commonly used in India. The lead concentrations listed here are from medicines manufactured in the United States. Assume that a simple random sample has been selected. Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 14.0 µg/g. 5.98 5.50 20.54 3.03 6.46 Identify the null and alternative hypotheses. Ho: H 14 H₁: μ 14 (Type integers or decimals. Do not round.) Identify the test statistic. = (Round to two decimal places as needed.) 7.45 12.01 20.47 11.48 17.53 D S Vi I. (1,0) More
Do cars really get better mileage per gallon on the highway? The table shows results from a study of the MPG (miles per gallon) of cars both in the city and on the highway. Assume that the two samples are randomly selected, independent, the population standard deviations are not know and not considered equal. At the 0.1 significance level, test the claim that the mpg on the highway is better than in the city. MPG on the Highway 35.6 34.3 32.2 33.9 31.1 27.1 33.3 33.4 29.3 33.5 31.4 33.2 33.5 30.8 33.8 MPG in the City 26.4 25.3 18.6 25.6 24.7 24.6 25.1 22.4 29.3 23.7 23.4 22 24 23.6 25.5 What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)H0: Select an answer p x̄₁ p₁ σ₁² μ₁ μ₂ μ μ(Highway) x̄₂ p̂₁ s₁² p₂ ? ≤ ≠ < ≥ = > Select an answer p₁ p₂ p̂₁ μ(City) μ μ₁ μ₂ x̄₁ x̄₂ s₁² σ₁² p H1: Select an answer p₂ μ(Highway) p̂₂ σ₂² x̄₁ x̄₂ s₂² μ₁ μ₂ μ p₁ p ? < ≠ = ≥ ≤ > Select an answer p₂ p₁ μ₁ σ₁²…
Given below are the analysis of variance results from a stat software display comparing sample data for the meanmileage for 4 different types of cars. Assume that you want to use a 0.05 significance level in testing the null hypothesisthat the different samples come from cars with the same mileage. What can you conclude about the equality of the population means?Source DF SS MS F pFactor 3 13.500 4.500 5.17 0.011Error 16 13.925 0.870Total 19 27.425 12)What is the null and alternative hypothesis for this scenari?Based on this One Way Anova software analysis, would you reject of fail to reject the nul hypothesis?What does this result mean in context?
Do men score lower on average compared to women on their statistics finals? Final exam scores of twelve randomly selected male statistics students and twelve randomly selected female statistics students are shown below. Male: 89 94 88 64 57 58 62 83 82 88 57 78 Female: 81 80 65 66 97 92 70 94 82 90 65 93 Assume both follow a Normal distribution. What can be concluded at the the a = 0.10 level of significance level of significance? For this study, we should use Select an answer a. The null and alternative hypotheses would be: Ho: Select an answer Select an answer Select an answer v (please enter a decimal) H1: Select an answer v Select an answer v Select an answer ♥ (Please enter a decimal) b. The test statistic ? v = (please show your answer to 3 decimal places.) c. The p-value = d. The p-value is ? v a e. Based on this, we should Select an answer v the null hypothesis. f. Thus, the final conclusion is that ... (Please show your answer to 4 decimal places.) O The results are…
Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. Identify the value of the test statistic. Source DF MS F Factor 3 13.500 4.500 5.17 0.011 Error 16 13.925 0.870 Total 19 27.425 Seleccione una: O A. 5.17 О В. 0.011 O C. 4.500 O D. 13.500
Start A sales manager for a large department store believes that customer spending per visit with a sale is lower than customer spending without a sale, and would like to test that claim. A simple random sample of customer spending is taken from without a sale and with a sale. The results are shown below. Without sale with sale 82.904 71.316 Mean Variance 1984.96 1826.41 Observations 150 Hypothesized Mean Difference 328 PT<=t) one-tail 0.0071 Critical one-tail -1.65 PT<=t) two-tail Critical two-tail -1.967 6 Confidence Level 3 Ex: 1.234 Samples from without sale: n₁ = Ex 9 Samples from with sale: n₂ Point estimate for spending without sale: 1 Point estimate for spending with sale: 2 = 0 J. Ex: 1.234 EN P -7 T P= = t= DE UNDE IS-

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