From the information below, interpret the skewness and kurtosis values and how you determined whether the assumption of normality was met or violated. Descriptive Statistics N Minimum Maximum Mean Std. Deviation Skewness Kurtosis Statistic Statistic Statistic Statistic Statistic Statistic Std. Error Statistic Std. Error quiz1 105 0 10 7.47 2.481 -0.851 0.236 0.162 0.467 gpa 105 1.08 4.00 2.8622 0.71266 -0.220 0.236 -0.688 0.467 total 105 54 123 100.09 13.427 -0.757 0.236 1.146 0.467 final 105 40 75 61.84 7.635 -0.341 0.236 -0.277 0.467 Valid N (listwise) 105
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A skewness value of 0 indicates a perfectly symmetrical distribution. Positive skewness indicates a distribution with an elongated tail on the right side, while negative skewness indicates a distribution with an elongated tail on the left side. - Kurtosis measures Show more…
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Descriptive Statistics for GPA and Quiz 3: Mean: 2.86 (GPA), 7.13 (Quiz 3) Standard Deviation: 0.71 (GPA), 1.60 (Quiz 3) Skewness: -0.22 (GPA), -0.08 (Quiz 3) Standard Error of Skewness: 0.24 (GPA), 0.24 (Quiz 3) Kurtosis: -0.69 (GPA), 0.15 (Quiz 3) Standard Error of Kurtosis: 0.47 (GPA), 0.47 (Quiz 3)
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Normality Consider the following data. The summary statistics, histogram, and normal quantile plot were generated using Minitab.(a) Does the histogram indicate normality for the data distribution? Explain. (b) Does the normal quantile plot indicate normality for the data distribution? Explain. (c) Compute the interquartile range and check for outliers. (d) Compute Pearson's index. Does the index value indicate skewness? (e) Using parts (a) through (d), would you say the data are from a normal distribution?
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Find the value of $t \neq$ for the difference between two means based on an assumption of normality and this information about two samples: $$\begin{array}{cccc}\text { Sample } & \text { Number } & \text { Mean } & \text { Std. Dev. } \\\hline 1 & 21 & 1.66 & 0.29 \\2 & 9 & 1.43 & 0.18 \\\hline\end{array}$$
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