Chapter 66: Problem 6
The resistance of a sample of 400 resistors produced by an automatic process have the following resistance distribution. \begin{tabular}{l} Resistance Frequency \\ \((\Omega)\) \\ \(50.11\) & 9 \\ \(50.15\) & 35 \\ \(50.19\) & 61 \\ \(50.23\) & 102 \\ \(50.27\) & 89 \\ \(50.31\) & 83 \\ \(50.35\) & 21 \\ \hline \end{tabular} Calculate for the sample: (a) the mean and (b) the standard deviation. (c) Test the null hypothesis that the resistances of the resistors are normally distributed at a level of significance of \(0.05\), and determine if the distribution gives a 'too good' fit at a level of confidence of \(90 \%\)
Short Answer
Step by step solution
Calculate the Mean
Finish Calculating the Mean
Calculate the Standard Deviation
Standard Deviation - Detailed Calculation
Finish Calculating the Standard Deviation
Test the Normality Hypothesis
Conclude Normality Test Result
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mean Calculation
Standard Deviation
- \( s = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{N-1}} \)
Normal Distribution Hypothesis
Chi-Squared Test
- \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \)