A one-way ANOVA was conducted to compare the means of four groups, each of which had 10 participants. What is the critical F for this analysis at the alpha = .05 level?
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- In a Completely Randomized Experimental Design, 3 brands of paper towels were tested for their ability to absorb water. Equal-sized towels were used, with three sections of towels tested per brand. Excel was used to perform the ANOVA analysis of the data using alpha = 0.05. The Excel-generated output is given below: SUMMARY Groups Count Sum Average Variance Brand X 3 280 93.33333333 34.33333333 Brand Y 3 289 96.33333333 6.333333333 Brand Z 3 248 82.66666667 42.33333333 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 309.5555556 2 154.7777778 5.59437751 0.042532458 5.14325285 Within Groups 166 6 27.66666667 Total 475.5555556 8 The null hypothesis states that there is no significant difference between the population mean absorbency ratings for these three different brands of paper towel. After looking at the Excel-generated ANOVA output above,…The work week for adults in the US that work full time is normally distributed with a mean of 47 hours. A newly hired engineer at a start-up company believes that employees at start-up companies work more on average then most working adults in the US. She asks 12 engineering friends at start-ups for the lengths in hours of their work week. Their responses are shown in the table below. Test the claim using a α=0.05 level of significance. Give answer to at least 4 decimal places. Hours 48 41 60 55 48 68 53 55 50 49 54 52 What are the correct hypotheses? H0: hours Ha: hoursThe result of an One-Way ANOVA for total stress symptoms among four groups of teachers (pre-k, elementary, middle, high school) was F(3, 117) = 3.827, p < .05. The corresponding effect size was η2 = .26. What can be concluded about these results?
- Before the furniture store began its ad campaign, it averaged 158 customers per day. The manager is investigating if the average is smaller since the ad came out. The data for the 11 randomly selected days since the ad campaign began is shown below: 147, 142, 149, 133, 146, 159, 132, 137, 148, 172, 150 Assuming that the distribution is normal, what can be concluded at the αα = 0.10 level of significance? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer > ≠ < = H1:H1: ? p μ Select an answer < ≠ > = The test statistic ? z t = (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 Select an answer accept fail to reject reject the null hypothesis. Thus, the final conclusion is that ... The data suggest…In a clinical trial, 16 out of 857 patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that 1.5% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 1.5% of this drug's users experience flulike symptoms as a side effect at the x = 0.05 level of significance? Because npo (1-Po) = 10, the sample size is 5% of the population size, and the sample ✓the requirements for testing the hypothesis ✓ satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? Ho: (Type integers or decimals. Do not round.) Find the test statistic, Zo. Zo = Find the P-value. P-value = (Round to three decimal places as needed.) Choose the correct conclusion below. A. Since P-value > α, reject the null hypothesis and conclude that there is not sufficient evidence that more than 1.5% of the users experience flulike symptoms. B. Since P-value α, do not reject the null…Facebook: A study showed that two years ago, the mean time spent per visit to Facebook was 20.8 minutes, Assume the standard devlation is o= 8.0 minutes. Suppose that a simple random sample of 107 visits was selected this year and has a sample mean of x= 18.8 minutes. A social scientist is interested to know whether the mean time of Facebook visits has decreased. Use the a=0.10 level of significance and the critical value method with the table. (a) State the appropriate null and alternate hypotheses. (b) Compute the value of the test statistic. (c) State a conclusion. Use the a = 0.10 level of significance.
- Suppose you are a product engineer for a tool manufacturing company, and want to investigate the strength of several new prototype redesigns of one of your drills. You have 15 new drill designs and would like to include a control group also of the original drill design. Your firm has decided to sample 33 drills for each new drill design and 33 drills in the control group with the original drill design. Using this information, you conduct an ANOVA test based on a quantitative measure of the drill's performance, and find that there is a sum of squares between treatments of 2594.0, along with a sum of squares error of 5245.8. Using this information, what was the F-statistic of your analysis? Only round final answer. Round to two decimal places.English Air continually monitors the proportion of overweight items checked by passengers on its flights in order to evaluate the appropriateness of their overweight fees. Recently, a random sample of 288 items checked on English Air flights to North America contained 51 overweight items, and an independent, random sample of 258 items checked on English Air flights within Europe contained 27 overweight items. Based on these samples, can we conclude, at the 0.05 level of significance, that there is a difference between the proportion p, of all items on English Air flights to North America that are overweight and the proportion p, of all items on English Air flights within Europe that are overweight? Perform a two-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 parts below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H, and the alternative hypothesis…Question The result of a One-Way ANOVA for total stress symptoms among four groups of teachers (pre-k, elementary, middle, high school) was F(3, 117) = 3.82, p < .05. The corresponding effect size was η2 = .265. What can be concluded about these results? Answer choices: For total stress symptoms, there was not significant difference among groups. For total stress symptoms, there was a significant difference among groups. The effect size was small. For total stress symptoms, there was a significant difference among groups. The effect size was medium. For total stress symptoms, there was a significant difference among groups. The effect size was very large.
- In a clinical trial, 25 out of 852 patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that 2.4% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 2.4% of this drug's users experience flulike symptoms as a side effect at the a = 0.05 level of significance? Because npo (1- Po) = 10, the sample size is 5% of the population size, and the sample the requirements for testing the hypothesis satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? Ho: versus H1: (Type integers or decimals. Do not round.)Suppose you are a product engineer for a tool manufacturing company, and want to investigate the strength of several new prototype redesigns of one of your drills. You have 13 new drill designs and would like to include a control group also of the original drill design. Your firm has decided to sample 37 drills for each new drill design and 37 drills in the control group with the original design. Using this information, if you were to conduct an ANOVA test based on a quantitative measure of the drill's performance, what is the degree of freedom associated with the variation within groups (SSE) in this analysis (denominator degree of freedom)?Before the furniture store began its ad campaign, it averaged 159 customers per day. The manager is investigating if the average has changed since the ad came out. The data for the 12 randomly selected days since the ad campaign began is shown below: 191, 161, 179, 160, 166, 181, 168, 163, 180, 164, 191, 147 Assuming that the distribution is normal, what can be concluded at the αα = 0.05 level of significance? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? p μ Select an answer = > < ≠ H1:H1: ? p μ Select an answer = ≠ > < The test statistic ? z t = (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 Select an answer fail to reject accept reject the null hypothesis. Thus, the final conclusion is that ... The…
