To conduct a goodness-of-fit test to determine whether a sample appears to be selected from a normal distribution, we can use a chi-square test. Let's set up the problem step by step, focusing on the intervals that define the normal distribution.
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Sample Parameters:
- Sample size \(n = 40\)
- Sample mean \(\bar{x} = 63\)
- Sample standard deviation \(s = 12\)
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Assuming Normal Distribution:
If the sample is from a normal distribution, we can compute the theoretical frequencies for different intervals based on the mean and standard deviation. -
Intervals in a Normal Distribution:
The normal distribution can be divided into intervals based on standard deviations from the mean. The intervals covering the data can be defined in terms of the mean and standard deviation as follows:- 1st interval: \((-\infty, \mu - 3\sigma)\)
- 2nd interval: \((\mu - 3\sigma, \mu - 2\sigma)\)
- 3rd interval: \((\mu - 2\sigma, \mu - \sigma)\)
- 4th interval: \((\mu - \sigma, \mu + \sigma)\)
- 5th interval: \((\mu + \sigma, \mu + 2\sigma)\)
- 6th interval: \((\mu + 2\sigma, \mu + 3\sigma)\)
- 7th interval: \((\mu + 3\sigma, \infty)\)
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Calculating the Boundaries:
With \(\mu = 63\) and \(\sigma = 12\):- \(\mu - 3\sigma = 63 - 3 \times 12 = 63 - 36 = 27\)
- \(\mu - 2\sigma = 63 - 2 \times 12 = 63 - 24 = 39\)
- \(\mu - \sigma = 63 - 12 = 51\)
- \(\mu + \sigma = 63 + 12 = 75\)
- \(\mu + 2\sigma = 63 + 2 \times 12 = 63 + 24 = 87\)
- \(\mu + 3\sigma = 63 + 3 \times 12 = 63 + 36 = 99\)
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Fourth Interval:
The fourth interval is from the mean minus one standard deviation to the mean plus one standard deviation.- The boundaries are: \[ (\mu - \sigma, \mu + \sigma) = (51, 75) \]
Thus, the boundary of the fourth interval is between 51 and 75.