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- la). The null hypothesis is that the difference in the grams of sugar between ultra-processed and processed food μ = 0 (Ho: μ = 0). Write an example alternative hypothesis for the mean (µ) difference between ultra-processed foods (up) and processed foods (μ). 1b). Next, generate 1,000 samples. Let's say that you asked students in our class to calculate the difference in the amount of sugar between their ultra-processed food and processed food at home, which had a mean (u) sugar difference of 14.75 grams of sugar. Do you have evidence to reject the null hypothesis? 1c). Let's say that your sample difference in the amount of sugar between ultra-processed food and processed food at home was 4.75 grams of sugar. Do you have evidence to reject the null hypothesis? 1d). What is a 92% confidence interval for the data? Please write it using the full confidence interval interpretation language (e.g., I am 92% confident that...). le). What percent of the data lie to the left of -14.25 grams of…Four machines M - Mawere each made with three different components B - Bg fitted. The average production capacities [kg / h] of the machines were determined for each component during a test phase. The following values were measured: machine 3 4th B 7.07 8.24 8.43 9.2 B2 9.1 7.98 8.87 8.11 6.07 7th 5.98 6.36 Determine whether there are significant differences in terms of components or machines and calculate the effect sizes, Normal distribution with the same variance can be assumed. Note: q = 14.9477, 1 = 11.0648, 9a =3.5125 Name of the test procedure: O One-way analysis of variance Ot-test for independent samples with homogeneous variances O Two-factor analysis of variance with repeated measures O Two-factor analysis of variance without repetition of measurements Significance level: O 0.5 O 0.05 O 0.01 Overall hypothesis H: O There are no significant differences in terms of components and machines O There are no significant differences in terms of components or machines O There are no…To compare the effectiveness of competing fertilizers on increasing radish size, researchers randomly divided 20 radish seeds of the same type into two groups of 10 and planted them. To one of the groups, they applied Fertilizer A, and to the other, they applied Fertilizer B. Then the researchers measured the radii of the radishes grown from the seeds in the two groups. The mean radius of the Fertilizer A group was 0.52 inches smaller than the mean radius of the Fertilizer B group. The researchers wanted to determine if this difference was due to the type of fertilizer or simply due to more seeds that produced smaller radishes being randomly assigned to the Fertilizer A group. To test if a difference of the means (A - B) of -0.52 inches or less could be due to chance, the researchers used a computer simulation to shuffle and randomly reassign the radishes to the two fertilizer groups. The difference between the two group means was then calculated and plotted on a dot plot. They…
- 1. Model 1: OLS, using observations 1-706 Dependent variable: RST Coefficient Std. Error t-ratio p-value const 3586.38 38.9124 92.17 <0.0001 *** TOTWRK −0.150746 0.0167403 −9.005 <0.0001 *** Mean dependent var 3266.356 S.D. dependent var 444.4134 Sum squared resid 1.25e+08 S.E. of regression 421.1357 R-squared 0.103287 Adjusted R-squared 0.102014 F(1, 704) 81.08987 P-value(F) 1.99e-1810538.19 Log-likelihood −5267.096 Akaike criterion 10538.19 Schwarz criterion 10547.31 Hannan-Quinn 10541.71 RSTi =3586.38−0.150746 x TOTWRKi , R2=0.103287,SER=421.1357 (38.9124) (0.0167403) Question? could you please help with this question below. 3) By observing the GRETL output in Part (1) above, provide a detailed explanation of the coefficient of determination. Based on your analysis, is this a good model? Why or why not?To compare the dry braking distances from 30 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 35 models of Make A and 35 models of Make B. The mean braking distance for Make A is 43 feet. Assume the population standard deviation is 4.6 feet. The mean braking distance for Make B is 46 feet. Assume the population standard deviation is 4.5 feet. At α=0.10, can the engineer support the claim that the mean braking distances are different for the two makes of automobiles? Assume the samples are random and independent, and the populations are normally distributed. The critical value(s) is/are Find the standardized test statistic z for μ1−μ2.1. If the F ratio in an ANOVA is more extreme than the cutoff score, what does this suggest? a. there is a significant difference between population 1 and population 2 b. there is a significant difference between population 2 and population 3 c. there is a significant difference between populations (but we don’t know where it is) d. there is insufficient evidence for the research hypothesis 2. The equation for the best fit line of a scatter plot (also called the linear prediction rule) is what? a. y = a + bxc. t = r / √((1-r2)/df) b. β = b*(√SSX/√SSY) d. a = (MY – (b)(MX))/df
- B6. Pinan Insurance Company wants to study the relationship between the amount of fire damage and the distance between the burning house and the nearest fire station. This information will be used in setting rates for insurance coverage. For a sample of 30 claims for the last year, the director of the actuarial department determined the distance from the fire station (x) and the amount of fire damage, in thousands of dollars (y). The MegaStat output is reported below. ANOVA table Source Regression Residual Total Regression output Variables Intercept Distance-X Answer the following questions. SS 1,864.5782 1,344.4934 3,209.07 Coefficients 12.3601 4.7956 df 1 28 29 MS 1,864.5782 48.0176 Std. Error 3.2915 0.7696 F 38.83 t(df=28) 3.755 6.231 a) Write out the regression equation. Is there a direct or indirect relationship between the distance from the fire station and the amount of fire damage? b) How much damage would you estimate for a fire 5 miles from the nearest fire station?Answer the following: 1. Test Stat 2. Critical Region: Select one: a. T1.761 c. T>2.14 d. T>abs(2.145) 3. Conclusion: At 5% level of significance… Select one: a. the new type of fabric does not exceed the WRA of the top current brand b. the new type of fabric exceeds the WRA of the top current brand34) If a population exhibits a skew of -5.00 and an excess kurtosis of 5.00 which of the following is true? The population mean exceeds the median. The population variance exceeds the standard deviation. The population standard deviation exceeds the variance. The population median equals the mean. The population median exceeds the mean.
- 10. 18.30 Temperatures are measured at various points on a heated plate (Table P18.30). Estimate the temperature at (a) x = 4, y = 3.2, and (b) x = 4.3, y = 2.7. TABLE P18.30 Temperature (°C) at various points on a square heated plate. x = 0 x = 2 y = 0 y 2 y = 4 y=6 y 8 100.00 85.00 70.00 55.00 40.00 90.00 64.49 48.90 38.78 35.00 x = 6 70.00 48.15 35.03 30.39 27.07 30.00 25.00 x = 4 80.00 53.50 38.43 x = 8 60.00 50.00 40.00 30.00 20.0020. Find the coefficient of skewness from the tollowing intormation : Difference of two quartiles = 8 Mode = 11 eboM %3D Sum of two quartiles = 22 Mean = 8 %3DConstruct a dot plot for the asphalt data.(In the article “Evaluation of Low-Temperature Properties of HMA Mixtures”, the following values of fracture stress (in megapascals) were measured for a sample of 24 mixtures of hot-mixed asphalt (HMA).30 75 79 80 80 105 126 138 149 179 179 191223 232 232 236 240 242 245 247 254 274 384 470