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Within the class, my classmates and I are given a fixed daily allowance by our parents. Curious, I wanted to know the average daily allowance of my fellow classmates. The daily allowances of the 15 students from my class are given below. Daily Allowance (in PHP) 1 20 2 50 3 150 4 50 5 50 6 50 IT 7 70 8 50 9 100 10 30 11 80 12 100 13 50 50 80 Student 14 15 What is the probability that a random sample of size 7 out of 15 identified students will have an average daily allowance of 50 or more?

Within the class, my classmates and I are given a fixed daily allowance by our parents. Curious, I wanted to know the average daily allowance of my fellow classmates. The daily allowances of the 15 students from my class are given below. Daily Allowance (in PHP) 1 20 2 50 3 150 4 50 5 50 6 50 IT 7 70 8 50 9 100 10 30 11 80 12 100 13 50 50 80 Student 14 15 What is the probability that a random sample of size 7 out of 15 identified students will have an average daily allowance of 50 or more?

Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter2: Working With Real Numbers
Section2.3: Rules For Addition
Problem 8P
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Within the class, my classmates and I are given a fixed daily allowance by our parents. Curious, I wanted to know the average daily allowance of my fellow classmates. The daily allowances of the 15 students from my class are given below. Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Mean SD Daily Allowance (in PHP) 20 50 150 50 50 50 70 50 100 30 80 100 50 50 80 Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Sample size Hypothesized Mean test statistic DF p-value Significance Level t-score Probability What is the probability that a random sample of size 7 out of 15 identified students will have an average daily allowance of 50 or more? Daily Allowance (in PHP) 20 50 150 50 50 50 70 50 100 30 80 100 50 50 80 Sample Mean 65.33333333 32.70357487 7 50 1.240481719 6 0.869448642 -1.94 1.24 5.00% 5.00% Because the p-value of our test (0.86) is greater than alpha = 0.05, we fail to reject the null hypothesis of the test. There is no sufficient evidence to say that the mean daily allowance for classmates is greater than 50. From the given data μ = 50 n=7 Null Hypothesis: μ ≥ 50 For computing the probability t-distribution is used as the sample size is < 30 t = 65.333-50 32.70357 √7 t = 1.24 Hence, the probability that 7 students' daily allowance of 50 or more = 0.87 For (n-1)=7-1= 6 degrees of freedom, the one-tailed test value for t -distribution = 1.94 at 5% level of significance Rejection Region DO NOT REJECT THE NULL HYPOTHESIS The sample means as extreme as 65.333 86.94% of the time under the null hypothesis. Null Hypothesis: μ ≥ 50 Alternative Hypothesis: μ<50 Since 86.94% is greater than the significance level of 5.00%, Hence there is no sufficient evidence to reject the null hypothesis. 20 Sample Mean Under the Null Distribution 40 60 80 100
Transcribed Image Text:Within the class, my classmates and I are given a fixed daily allowance by our parents. Curious, I wanted to know the average daily allowance of my fellow classmates. The daily allowances of the 15 students from my class are given below. Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Mean SD Daily Allowance (in PHP) 20 50 150 50 50 50 70 50 100 30 80 100 50 50 80 Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Sample size Hypothesized Mean test statistic DF p-value Significance Level t-score Probability What is the probability that a random sample of size 7 out of 15 identified students will have an average daily allowance of 50 or more? Daily Allowance (in PHP) 20 50 150 50 50 50 70 50 100 30 80 100 50 50 80 Sample Mean 65.33333333 32.70357487 7 50 1.240481719 6 0.869448642 -1.94 1.24 5.00% 5.00% Because the p-value of our test (0.86) is greater than alpha = 0.05, we fail to reject the null hypothesis of the test. There is no sufficient evidence to say that the mean daily allowance for classmates is greater than 50. From the given data μ = 50 n=7 Null Hypothesis: μ ≥ 50 For computing the probability t-distribution is used as the sample size is < 30 t = 65.333-50 32.70357 √7 t = 1.24 Hence, the probability that 7 students' daily allowance of 50 or more = 0.87 For (n-1)=7-1= 6 degrees of freedom, the one-tailed test value for t -distribution = 1.94 at 5% level of significance Rejection Region DO NOT REJECT THE NULL HYPOTHESIS The sample means as extreme as 65.333 86.94% of the time under the null hypothesis. Null Hypothesis: μ ≥ 50 Alternative Hypothesis: μ<50 Since 86.94% is greater than the significance level of 5.00%, Hence there is no sufficient evidence to reject the null hypothesis. 20 Sample Mean Under the Null Distribution 40 60 80 100
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