Are any of these results significant? ### Analysis of Variance (ANOVA) - Intended Screen TimeThis table presents an ANOVA summary for the analysis of intended screen time. The purpose of this analysis is to determine if there are statistically significant differences between groups based on the variables analyzed. Below is a detailed explanation of each component of the table:| Source | Sum of Squares | df | Mean Square | F | p | η² ||-------------------------|----------------|----|-------------|---------|-------|------|| **Reference Group** | 0.0307 | 1 | 0.0307 | 0.0216 | 0.884 | 0.000 || **Norm** | 3.4376 | 1 | 3.4376 | 2.4202 | 0.124 | 0.029 || **Reference Group ⨯ Norm** | 1.7006 | 1 | 1.7006 | 1.1973 | 0.277 | 0.014 || **Residuals** | 112.2100 | 79 | 1.4204 | | | |#### Explanation of Terms:- **Sum of Squares**: This represents the total variation attributed to each factor.- **df (Degrees of Freedom)**: This indicates the number of values that are free to vary for each source of variation.- **Mean Square**: Calculated as the sum of squares divided by the degrees of freedom, indicating the average variation.- **F**: The F-statistic, which is used to assess whether there are any significant differences between the means of the groups.- **p (p-value)**: This value indicates the probability of observing the data, assuming that the null hypothesis is true. A lower p-value (typically < 0.05) suggests significant differences.- **η² (Eta Squared)**: This is a measure of the effect size, indicating the proportion of the total variance that is attributed to an effect.#### Interpretation:- The **Reference Group** factor contributes minimally to the variation in intended screen time as indicated by the very low F-value and non-significant p-value (0.884).- The **Norm** factor shows a larger contribution to variance, but it is not statistically significant with a p-value of 0

The ANOVA table for Intended Screen Time is given.

Answered: Are any of these results significant?

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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A recent issue of the AARP Bulletin reported that the average weekly pay for a woman with a high school degree is $520 . Suppose you would like to determine if the average weekly pay for all working women is signicantly greater (at alpha = 0.05) than that for women with a high school degree. Using data on weekly pay for a sample of 50 working women, you obtain the following statistics: sample mean = $620, sample error = $150. Conduct the hypothesis test using the critical value approach and the p-value approach.Describe the conclusion of the hypothesis test and what that means in this context
The authors of a paper classified characters who were depicted smoking in movies released between a certain range of years. The smoking characters were classified according to sex and whether the character type was positive, negative, or neutral. The resulting data are summarized in the accompanying table. Assume that it is reasonable to consider this sample of smoking movie characters as representative of smoking movie characters. Do the data provide evidence an association between sex and character type for movie characters who smoke? Use α = 0.05. Sex = Male Female Positive Negative Neutral 256 Character Type 85 107 11 132 51 Calculate the test statistic. (Round your answer to two decimal places.) x² Use technology to calculate the P-value. (Round your answer to four decimal places.) P-value = What can you conclude? O Reject Ho. There is convincing evidence to conclude that there is an association between sex and character type for movie characters who smoke. Fail to reject Ho.…
QUESTION 4 PART A- What is the standard error of the correlation coefficient for the association of extroversion and neuroticism in personalities? QUESTION 4 PART B- Carry out a formal t-test for the association between extroversion and neuroticism in personalities.xls. H0: The population correlation coefficient is 0. HA: The population correlation coefficient is not 0. What is the value of the t-statistic corresponding to this test? (Report the absolute value. The t-distribution is symmetric, thus it is easier to compare the absolute value of the t-statistic with the positive values in statistical table C). DATA IS ATTACHED.
Managers of an outdoor coffee stand in Coast City are examining the relationship between (hot) coffee sales and daily temperature, hoping to be able to predict a day's total coffee sales from the maximum temperature that day. The bivariate data values for the coffee sales (denoted by y , in dollars) and the maximum temperature (denoted by x , in degrees Fahrenheit) for each of fifteen randomly selected days during the past year are given below. These data are plotted in the scatter plot in Figure 1. Temperature, x(in degrees Fahrenheit) Coffee sales, y(in dollars) 53.1 2215.5 76.6 1982.7 70.1 1937.3 39.7 2251.1 72.5 1603.9 48.4 2024.2 46.7 2135.3 83.9 1536.5 58.4 1965.6 38.5 1944.3 68.2 1746.1 75.4 1472.1 65.6 1819.5 53.9 1627.4 45.1 1782.3 Send data to calculator Send data to Excel Coffee sales(in dollars) y 1200 1400 1600 1800 2000 2200 2400 x 40 50 60 70 80 90…
Question 1. Greater than or Less than. Question 2. IS or IS NOT.
Managers of an outdoor coffee stand in Coast City are examining the relationship between (hot) coffee sales and daily temperature, hoping to be able to predict a day's total coffee sales from the maximum temperature that day. The bivariate data values for the coffee sales (denoted by y, in dollars) and the maximum temperature (denoted by x, in degrees Fahrenheit) for each of fifteen randomly selected days during the past year are given below. These data are plotted in the scatter plot in Figure 1. Also given is the product of the temperature and the coffee sales for each of the fifteen days. (These products, written in the column labelled "xy", may aid in calculations.) Temperature, X 10 Coffee sales, y (in dollars) (in degrees Fahrenheit) 37.2 51.3 59.7 53.7 63.1 44.5 47.5 75.9 82.2 40.0 74.8 70.6 69.0 48.3 74.7 Send data to calculator V 2000.4 2205.9 1944.8 1579.3 1846.3 1797.3 2016.5 1563.3 1556.5 2249.4 1653.9 1923.0 1747.9 2154.6 1979.7 74,414.88 113,162.67 116,104.56 84,808.41…
Managers of an outdoor coffee stand in Coast City are examining the relationship between (hot) coffee sales and daily temperature, hoping to be able to predict a day's total coffee sales from the maximum temperature that day. The bivariate data values for the coffee sales (denoted by y, in dollars) and the maximum temperature (denoted by x, in degrees Fahrenheit) for each of sixteen randomly selected days during the past year are given below. These data are plotted in the scatter plot in Figure 1. Temperature, x (in degrees Fahrenheit) Coffee sales, y (in dollars) 71.5 1970.9 58.7 1953.9 53.7 1791.1 2400+ 83.0 1570.3 2200+ 62.8 1852.7 74.6 1633.6 2000- ** 40.4 1973.9 1800- 51.5 2250.9 44.6 1808.5 1600- 45.5 1977.3 1400- 45.6 2190.5 1200 69.4 1789.1 55.0 1598.2 50 70 80 90 40.6 2272.0 74.0 1937.0 Figure 1 76.6 1547.1 Send data to Excel Continue Submit Assignment O 2021 McGraw-Hill Education. All Rights Reserved. Terms of Use Privacy Accessibility here to search 99+ 5 4.
Question 2 Is the proportion of wildfires caused by humans in the south different from the proportion of wildfires caused by humans in the west? 380 of the 550 randomly selected wildfires looked at in the south were caused by humans while 404 of the 572 randomly selected wildfires looked at the west were caused by humans. What can be concluded at the a = 0.01 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: Select an answer Select an answer Select an answer v (please enter a decimal) H: Select an answer v Select an answer V Select an answer |(Please enter a decimal) c. The test statistic (please show your answer to 3 decimal places.) d. The p-value = e. The p-value is ? a f. Based on this, we should Select an answer g. Thus, the final conclusion is that ... (Please show your answer to 4 decimal places.) the null hypothesis. O The results are statistically insignificant at a = 0.01, so there is insufficient…
Question 2 There is a claim that drink driving has been very common in high ways. The Highway Police randomly caught 300 drivers for breath alcohol test and found that 14% of drivers have breath alcohol over the legal limit. At 4% level of significance, use the p-value approach to test the hypothesis that the percentage of drivers who are drink driving is more than 10%.
Give answer from the choice and explain it as given in image. Note:hand written solution not allowed.
Question 2 A researcher was interested in studying if there is a significant relationship between the severity of COVID 19 and blood types of individuals. 2400 individuals were studied and the results are shown below. Condition Blood Type Blood Type Blood Type Blood Type Blood Type O A B AB Total Critical 64.00 44 20 8 136 Severe 175.00 129 50 15 369 Moderate 211.00 528 151 125 1015 Mild 200.00 400 140 140 880 Total 650.00 1101 361 288 2400 State both the null and alternative hypotheses. Provide the decision rule for making this decision. Use an alpha level of 5%.

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