Homework Help is Here – Start Your Trial Now!
Math
Probability
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?

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.1: Measures Of Center
Problem 4GP
See similar textbooks
icon
Related questions
Question

Answer the questions with complete SOLUTION, GRAPH & ANSWERS, just like in the given example

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
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Paths and Circuits
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, probability and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Glencoe Algebra 1, Student Edition, 9780079039897…
Glencoe Algebra 1, Student Edition, 9780079039897…
Algebra
ISBN:
9780079039897
Author:
Carter
Publisher:
McGraw Hill
SEE MORE TEXTBOOKS