Stat homework 1

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San Antonio College *

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1203

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Statistics

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Feb 20, 2024

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docx

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Uploaded by DukeFang13253

Statistics Math Sheet Question 1: Descriptive Statistics a) Calculate the mean, median, and mode for the following data set: 10, 15, 18, 20, 22, 22, 25, 30, 35, 40. b) Calculate the range and standard deviation for the same data set. Answer 1: a) Mean = (10 + 15 + 18 + 20 + 22 + 22 + 25 + 30 + 35 + 40) / 10 = 24.7 Median = (22 + 22) / 2 = 22 Mode = 22 b) Range = 40 - 10 = 30 Standard Deviation: Calculate the mean deviation for each data point: Mean Deviation = |X - Mean| Mean Deviation: 10: |10 - 24.7| = 14.7 15: |15 - 24.7| = 9.7 18: |18 - 24.7| = 6.7 20: |20 - 24.7| = 4.7

22: |22 - 24.7| = 2.7 22: |22 - 24.7| = 2.7 25: |25 - 24.7| = 0.3 30: |30 - 24.7| = 5.3 35: |35 - 24.7| = 10.3 40: |40 - 24.7| = 15.3 Square each mean deviation: Mean Deviation Squared: 14.7^2 = 216.09 9.7^2 = 94.09 6.7^2 = 44.89 4.7^2 = 22.09 2.7^2 = 7.29 2.7^2 = 7.29 0.3^2 = 0.09 5.3^2 = 28.09 10.3^2 = 106.09 15.3^2 = 234.09 Find the mean of the squared deviations: Mean of Squared Deviations = (216.09 + 94.09 + 44.89 + 22.09 + 7.29 + 7.29 + 0.09 + 28.09 + 106.09 + 234.09) / 10 = 76.21 Take the square root of the mean of the squared deviations:

Standard Deviation = √76.21 ≈ 8.73 Question 2: Probability Distribution Consider a fair six-sided die. What is the probability of rolling: a) A prime number? b) An even number? c) A number greater than 3? Answer 2: a) Prime numbers on a six-sided die are 2, 3, and 5. Thus, the probability of rolling a prime number is 3/6 or 1/2. b) Even numbers on a six-sided die are 2, 4, and 6. Thus, the probability of rolling an even number is 3/6 or 1/2. c) Numbers greater than 3 on a six-sided die are 4, 5, and 6. Thus, the probability of rolling a number greater than 3 is 3/6 or 1/2. Question 3: Hypothesis Testing Suppose the average height of students in a class is claimed to be 65 inches. A random sample of 25 students is taken, and their average height is found to be 63 inches with a standard deviation of 4 inches. At 5% level of significance, test the claim. Answer 3:

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Given: Population mean (μ) = 65 inches Sample mean ( ) = 63 inches x̄ Sample standard deviation (s) = 4 inches Sample size (n) = 25 Level of significance (α) = 0.05 Null hypothesis (H0): μ = 65 Alternative hypothesis (H1): μ ≠ 65 Calculate the test statistic (z-score): z = ( - μ) / (s / √n) x̄ = (63 - 65) / (4 / √25) = -2 / (4 / 5) = -2 / 0.8 = -2.5 Since this is a two-tailed test, the critical z-value for a 5% level of significance is ±1.96. Since |-2.5| > 1.96, we reject the null hypothesis. Conclusion: There is sufficient evidence to reject the claim that the average height of students in the class is 65 inches at the 5% level of significance.