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a study reports the following final notation f 3 32 9 50 p 05 how many total participants were involved in this study grou

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A study reports the following final notation: F (3, 32) = 9.50, p < .05. How many total participants were involved in this study? Group of answer choices 34 32 36

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Answer to a math question A study reports the following final notation: F (3, 32) = 9.50, p < .05. How many total participants were involved in this study? Group of answer choices 34 32 36

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To determine the total number of participants involved in this study, we need to look at the notation provided: F(3, 32) = 9.50.

Here, the numbers inside the parentheses represent the degrees of freedom for the numerator and denominator, respectively. The degrees of freedom for the numerator (df1) is 3, and the degrees of freedom for the denominator (df2) is 32.

The degrees of freedom for the numerator represent the number of groups being compared, and the degrees of freedom for the denominator represent the total number of participants minus the number of groups being compared.

To calculate the total number of participants, we need to add the degrees of freedom for the numerator (df1) and the degrees of freedom for the denominator (df2), and then subtract 1 (since the degrees of freedom include the number of groups being compared).

Total participants = df1 + df2 - 1

Total participants = 3 + 32 - 1 = 34

Therefore, the total number of participants involved in this study is \boxed{34}.

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Sodium 38.15 38.78 38.5 38.65 38.79 38.89 38.57 38.59 38.59 38.8 38.63 38.43 38.56 38.46 38.79 38.42 38.74 39.12 38.5 38.42 38.57 38.37 38.71 38.71 38.4 38.56 38.39 38.34 39.04 38.8 A supplier of bottled mineral water claims that his supply of water has an average sodium content of 36.6 mg/L. The boxplot below is of the sodium contents levels taken from a random sample of 30 bottles. With this data investigate the claim using SPSS to apply the appropriate test. Download the data and transfer it into SPSS. Check that your data transfer has been successful by obtaining the Std. Error of the mean for your data which should appear in SPSS output as 0.03900.. If you do not have this exact value, then you may have not transferred your data from the Excel file to SPSS correctly. Do not continue with the test until your value agrees as otherwise you may not have correct answers. Unless otherwise directed you should report all numeric values to the accuracy displayed in the SPSS output that is supplied when your data has been transferred correctly. In the following questions, all statistical tests should be carried out at the 0.05 significance level. Sample mean and median Complete the following concerning the mean and median of the data. mean = mg/L 95% CI: to mg/L Based upon the 95% confidence interval, is it plausible that the average sodium content is 36.9 mg/L? median: mg/L The median value is 36.9 mg/L. Skewness Complete the following concerning the skewness of the data. Skewness statistic = Std. Error = The absolute value of the skewness statistic less than 2 x Std. Error Therefore the data can be considered to come from a population that is . Normality test Complete the following summary concerning the formal testing of the normality of the data. H0: The data come from a population that normal H1: The data come from a population that normal Application of the Shapiro-Wilk test indicated that the normality assumption reasonable for sodium content (S-W( )= , p= ). Main test Using the guidelines you have been taught that consider sample size, skewness and normality, choose and report the appropriate main test from the following ( Appropriate ONE ) You have selected that you wish to report the one-sample t-test. H0: The mean sodium content equal to 36.9 mg/L H1: The mean sodium content equal to 36.9 mg/L Application of the one-sample t-test indicated that the mean is 36.9 mg/L (t( ) = , p = ). You have selected that you wish to report the Wilcoxon signed rank test. H0: The median sodium content equal to 36.9 mg/L H1: The median sodium content equal to 36.9 mg/L Application of the Wilcoxon signed rank test indicated that the median is 36.9 mg/L (z = , N = , p = ).