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Sagot :
Sure, let's go through the solution step-by-step:
### Part (a): Hypothesis Test
Given data:
- Sample size, [tex]\( n = 15 \)[/tex]
- Sample standard deviation, [tex]\( s = 5 \)[/tex]
- Null hypothesis standard deviation, [tex]\( \sigma_0 = 6 \)[/tex]
- Significance level, [tex]\( \alpha = 0.025 \)[/tex]
Step 1: State the null and alternative hypotheses
[tex]\[ H_0: \sigma = 6 \][/tex]
[tex]\[ H_a: \sigma < 6 \][/tex]
Step 2: Determine the test statistic
The test statistic for a chi-square test of a single standard deviation is given by:
[tex]\[ \chi^2 = \frac{(n-1) \cdot s^2}{\sigma_0^2} \][/tex]
Plugging in the values:
[tex]\[ \chi^2 = \frac{(15-1) \cdot 5^2}{6^2} = \frac{14 \cdot 25}{36} = \frac{350}{36} = 9.722 \][/tex]
So, the test statistic is [tex]\( \chi^2 = 9.722 \)[/tex].
Step 3: Make the decision
To make a decision, we would compare this chi-square statistic to the critical value from the chi-square distribution table for [tex]\( n-1 = 14 \)[/tex] degrees of freedom at [tex]\( \alpha = 0.025 \)[/tex].
However, since the test statistic [tex]\( \chi^2 = 9.722 \)[/tex] is given and it would be compared with the critical value from the table, we conclude:
[tex]\[ \boxed{9.722} \][/tex]
### Part (b): Confidence Interval
We are asked to find the 95% confidence interval for the population standard deviation.
Step 1: Determine the chi-square critical values
For a 95% confidence interval with [tex]\( n-1 = 14 \)[/tex] degrees of freedom, we need the chi-square critical values that correspond to the lower and upper tails of [tex]\( 2.5\% \)[/tex] each (since [tex]\( 1 - 0.95 = 0.05 \)[/tex] and each tail will take half the significance level).
Step 2: Calculate the confidence interval for variance
The confidence interval for the population variance [tex]\( \sigma^2 \)[/tex] is given by:
[tex]\[ \left( \frac{(n-1) \cdot s^2}{\chi^2_{\text{upper}}}, \frac{(n-1) \cdot s^2}{\chi^2_{\text{lower}}} \right) \][/tex]
Step 3: Convert to standard deviation
The confidence interval for the population standard deviation [tex]\( \sigma \)[/tex] is found by taking the square root of the variance limits:
[tex]\[ ( \sigma_{\text{lower}}, \sigma_{\text{upper}} ) \][/tex]
From the given computations, the variance limits were found first and then the square roots were taken:
[tex]\[ \sigma_{\text{lower}} = \sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{\text{upper}}}} = 3.6606 \][/tex]
[tex]\[ \sigma_{\text{upper}} = \sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{\text{lower}}}} = 7.8855 \][/tex]
Thus, the 95% confidence interval for the population standard deviation is:
[tex]\[ \boxed{(3.661, 7.886)} \][/tex]
Here we round the final values to three decimal places as mentioned. Therefore, the confidence interval is between [tex]\( 3.661 \)[/tex] and [tex]\( 7.886 \)[/tex].
These are the final results based on the provided data and calculations.
### Part (a): Hypothesis Test
Given data:
- Sample size, [tex]\( n = 15 \)[/tex]
- Sample standard deviation, [tex]\( s = 5 \)[/tex]
- Null hypothesis standard deviation, [tex]\( \sigma_0 = 6 \)[/tex]
- Significance level, [tex]\( \alpha = 0.025 \)[/tex]
Step 1: State the null and alternative hypotheses
[tex]\[ H_0: \sigma = 6 \][/tex]
[tex]\[ H_a: \sigma < 6 \][/tex]
Step 2: Determine the test statistic
The test statistic for a chi-square test of a single standard deviation is given by:
[tex]\[ \chi^2 = \frac{(n-1) \cdot s^2}{\sigma_0^2} \][/tex]
Plugging in the values:
[tex]\[ \chi^2 = \frac{(15-1) \cdot 5^2}{6^2} = \frac{14 \cdot 25}{36} = \frac{350}{36} = 9.722 \][/tex]
So, the test statistic is [tex]\( \chi^2 = 9.722 \)[/tex].
Step 3: Make the decision
To make a decision, we would compare this chi-square statistic to the critical value from the chi-square distribution table for [tex]\( n-1 = 14 \)[/tex] degrees of freedom at [tex]\( \alpha = 0.025 \)[/tex].
However, since the test statistic [tex]\( \chi^2 = 9.722 \)[/tex] is given and it would be compared with the critical value from the table, we conclude:
[tex]\[ \boxed{9.722} \][/tex]
### Part (b): Confidence Interval
We are asked to find the 95% confidence interval for the population standard deviation.
Step 1: Determine the chi-square critical values
For a 95% confidence interval with [tex]\( n-1 = 14 \)[/tex] degrees of freedom, we need the chi-square critical values that correspond to the lower and upper tails of [tex]\( 2.5\% \)[/tex] each (since [tex]\( 1 - 0.95 = 0.05 \)[/tex] and each tail will take half the significance level).
Step 2: Calculate the confidence interval for variance
The confidence interval for the population variance [tex]\( \sigma^2 \)[/tex] is given by:
[tex]\[ \left( \frac{(n-1) \cdot s^2}{\chi^2_{\text{upper}}}, \frac{(n-1) \cdot s^2}{\chi^2_{\text{lower}}} \right) \][/tex]
Step 3: Convert to standard deviation
The confidence interval for the population standard deviation [tex]\( \sigma \)[/tex] is found by taking the square root of the variance limits:
[tex]\[ ( \sigma_{\text{lower}}, \sigma_{\text{upper}} ) \][/tex]
From the given computations, the variance limits were found first and then the square roots were taken:
[tex]\[ \sigma_{\text{lower}} = \sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{\text{upper}}}} = 3.6606 \][/tex]
[tex]\[ \sigma_{\text{upper}} = \sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{\text{lower}}}} = 7.8855 \][/tex]
Thus, the 95% confidence interval for the population standard deviation is:
[tex]\[ \boxed{(3.661, 7.886)} \][/tex]
Here we round the final values to three decimal places as mentioned. Therefore, the confidence interval is between [tex]\( 3.661 \)[/tex] and [tex]\( 7.886 \)[/tex].
These are the final results based on the provided data and calculations.
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