QUESTION 24 Calculate the Mean Squared Between treaments value for a repeated measures ANOVA with 10 observations in each of the four conditions, given 5-Between=879.85 and SS-Within = 692.11." OMS-Within = 19.23 O MS-Between = 219.96 MS-Between = 230.70 MS-Between = 293.28
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- Suppose the total weight of passengers and luggage on an airline flight is normally distributed with mean 8914 and standard deviation 32 What is the 2.28th percentile of total passenger weight28. "Calculate the Mean Squared Between treaments value for a repeated measures ANOVA with 6 observations in each of the four conditions, given SS-Between = 888.73 and SS-Within = 635.23." MS-Within = 31.76 MS-Between = 222.18 MS-Between = 211.74 MS-Between = 296.2425
- 25An investigation of the relationship between traffic flow (measured in cars per day) and lead content (measured in micrograms per gram of dry weight) in the bark of trees near the highway yielded the following summary statistics: Sample Mean Sample Standard Deviation Lead Content (micrograms per gram dry weight) y¯=y¯= 680 sysy = 240 Traffic Flow (cars / day) x¯=x¯= 1750 sxsx = 800 The correlation between lead content and traffic flow was found to be r = 0.6 and a scatterplot showed the form to be linear. The least squares regression line for using X to predict Y was found to be y^=365+0.18xy^=365+0.18x. Which one of the following is a proper interpretation of the slope? Group of answer choices For each 1 car per day increase in traffic flow, we predict an corresponding increase of 0.18 in lead content of surrounding trees For each 0.6 car per day increase in traffic flow, we predict an corresponding increase of 365.18 in lead content of surrounding trees For each 1…An investigation of the relationship between traffic flow (measured in cars per day) and lead content (measured in micrograms per gram of dry weight) in the bark of trees near the highway yielded the following summary statistics: Sample Mean Sample Standard Deviation Lead Content (micrograms per gram dry weight) y¯=y¯= 680 sysy = 240 Traffic Flow (cars / day) x¯=x¯= 1750 sxsx = 800 The correlation between lead content and traffic flow was found to be r = 0.6 and a scatterplot showed the form to be linear. The least squares regression line for using X to predict Y was found to be y^=365+0.18xy^=365+0.18x. For trees in a given area, if the traffic flow is 14,000 cars per week, predict the lead content in their bark. Note: the traffic flow given in this question is NOT in the same units as the values used to calculate the line. Given value of x = 14,000 cars per week equals how many cars per day Predicted Value of y is how many micrograms per gram of dry weight
- An investigation of the relationship between traffic flow (measured in cars per day) and lead content (measured in micrograms per gram of dry weight) in the bark of trees near the highway yielded the following summary statistics: Sample Mean Sample Standard Deviation Lead Content (micrograms per gram dry weight) y¯=y¯= 680 sysy = 240 Traffic Flow (cars / day) x¯=x¯= 1750 sxsx = 800 The correlation between lead content and traffic flow was found to be r = 0.6 and a scatterplot showed the form to be linear. Since Traffic flow is the X-variable, its mean and standard deviation have been labelled x¯ and sxx¯ and sx. Since Lead content is the Y-variable, its mean and standard deviation have been labelled y¯ and syy¯ and sy . Determine slope and y-intercept of the least squares regression line for using X to predict Y. Group of answer choices Slope y-interceptThe mean age when people got married for the first time in the year 2000 was 26 years old with a population standard deviation of 6 years. A researcher thinks that marriage age has significantly changed since then. A survey of married persons of this generation was done to see if the mean age has changed. The sample of 50 married persons found that their mean age of first marriage was 27.7 years old. Do the data support the claim at the 10% significance level? What are the correct hypotheses? H0: yearsHa: years Based on the hypotheses, find the following: Test Statistic z = (Give answer to at least 4 decimal places) P-Value= (Give answer to at least 4 decimal places)26
- eights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches. Men the same age have mean height 69.3 inches with standard deviation 2.8 inches. (a) What is the z-score for a woman 77 inches tall? Z-Score = What is the z-score for a man 81 inches tall? Z-Score =The patient recovery time from a particular surgical procedure is normallu distrubuted with a mean of 5.3 and a standard deviation of 2.1 days. what is the median recovery time? what is the z-score for a patient who takes ten days to recover?Dr. Gomez conducted an experiment on studying alone vs. in a group. Twelve participants were randomly assigned to study a short chapter of a statistics test with either a group of students or alone. Twenty-four hours after the study session ended, the participants took an examination on the information in the chapter and the number of correct answers was recorded. Data from the experiment and the mean (X-bar) and standard deviation (SD) for each group are given below. Compute the difference between the means as the Alone minus Group Study Method. Study Method Alone (A) Group (G) X-barA = 445/6 = 74.167 SDA = 7.305 74 55 67 67 X-barG = 343/6 = 57.167 SDG= 6.646 87 55 77 47 68 59 72…