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2. The average weight of 9 Grade 6 pupils is 36 kg with a standard deviation of 0.4 kg. Is there a reason to believe that the pupil's average weight is less than 36.3 kg at 5% level of significance?

2. The average weight of 9 Grade 6 pupils is 36 kg with a standard deviation of 0.4 kg. Is there a reason to believe that the pupil's average weight is less than 36.3 kg at 5% level of significance?

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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2. The average weight of 9 Grade 6 pupils is 36 kg with a standard deviation of 0.4 kg. Is there a reason to believe that the pupil's average weight is less than 36.3 kg at 5% level of significance?
Transcribed Image Text:2. The average weight of 9 Grade 6 pupils is 36 kg with a standard deviation of 0.4 kg. Is there a reason to believe that the pupil's average weight is less than 36.3 kg at 5% level of significance?
Solution: It is a one-tailed test since the alternative hypothesis uses the symbol for greater than (>). And since o is known and n 2 30, we will use z-test. The level of significance is 0.05. From Table 1, the z-critical value is 1.645. Thus, we have: 71.5 – 70 1.5 z = 8 0.8 V100 z = 1.875 1.645 z-critical value The computed z-value is 1.875 which is greater than the critical value of 1.645. Therefore, we reject the null hypothesis and support and accept the alternative hypothesis that the population mean is greater than 70. Example 2: Consider the following given information. Use a = 0.01. Interpret the result. Null Hypothesis Alternative Hypothesis I = 124.5 Ho: µ = 127 Ha: μ < 127 µ = 127 n = 12 s = 5
Transcribed Image Text:Solution: It is a one-tailed test since the alternative hypothesis uses the symbol for greater than (>). And since o is known and n 2 30, we will use z-test. The level of significance is 0.05. From Table 1, the z-critical value is 1.645. Thus, we have: 71.5 – 70 1.5 z = 8 0.8 V100 z = 1.875 1.645 z-critical value The computed z-value is 1.875 which is greater than the critical value of 1.645. Therefore, we reject the null hypothesis and support and accept the alternative hypothesis that the population mean is greater than 70. Example 2: Consider the following given information. Use a = 0.01. Interpret the result. Null Hypothesis Alternative Hypothesis I = 124.5 Ho: µ = 127 Ha: μ < 127 µ = 127 n = 12 s = 5
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