Math / Data & Statistics

QuestionIQ scores: Scores on an IQ test are normally distributed. A sample of 12 IQ scores had standard deviation $s=6$ . (a) Construct an $80 \%$ confidence interval for the population standard deviation $\sigma$ . Round the answers to at least two decimal places. (b) The developer of the test claims that the population standard deviation is $\sigma=6$ . Does this confidence interval contradict this claim? Explain.
Part: $0 / 2$
Part 1 of 2
An $80 \%$ confidence interval for the population standard deviation is $\square<\sigma<\square$ .

Studdy Solution

STEP 1

What is this asking? We're trying to find a range of likely values for the true standard deviation of IQ scores, given a sample of 12 scores with a standard deviation of 6, and then see if the claimed standard deviation of 6 falls within that range. Watch out! Don't mix up standard deviation ( $\sigma$ ) with variance ( $\sigma^2$ ).
Also, remember that the chi-squared distribution is not symmetric, so we need to find two different critical values.

STEP 2

1. Calculate Degrees of Freedom
2. Find Critical Chi-Squared Values
3. Calculate the Lower Bound of the Confidence Interval
4. Calculate the Upper Bound of the Confidence Interval
5. Check the Claim

STEP 3

We start by figuring out the degrees of freedom, which is related to the sample size.
Since our sample size is $\text{n} = 12$ , the degrees of freedom is $\text{n}-1 = 12 - 1 = \bold{11}$ .
Remember, degrees of freedom basically tells us how much information we're getting from our sample!

STEP 4

For an 80% confidence interval, we have $\alpha = 1 - 0.80 = 0.20$ , so $\alpha/2 = 0.20/2 = \bold{0.10}$ .
We need to find two chi-squared values: $\chi^2_{1-\alpha/2}$ and $\chi^2_{\alpha/2}$ .

STEP 5

With 11 degrees of freedom and $\alpha/2 = 0.10$ , we look up these values in a chi-squared table or use a calculator.
We find $\chi^2_{1-0.10} = \chi^2_{0.90} = \bold{5.578}$ and $\chi^2_{0.10} = \bold{17.275}$ .
These values mark the boundaries of the middle 80% of the chi-squared distribution.

STEP 6

The formula for the lower bound is $\sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{\alpha/2}}}$ .
Plugging in our values, we get [\sqrt{\frac{(12-1) \cdot 6^2}{17.275}} = \sqrt{\frac{11 \cdot 36}{17.275}} = \sqrt{\frac{396}{17.275}} \approx \sqrt{22.91} \approx \bold{4.79}\).

STEP 7

The formula for the upper bound is \[\sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{1-\alpha/2}}}.
Plugging in our values, we get [\sqrt{\frac{(12-1) \cdot 6^2}{5.578}} = \sqrt{\frac{11 \cdot 36}{5.578}} = \sqrt{\frac{396}{5.578}} \approx \sqrt{71.00} \approx \bold{8.43}\).

STEP 8

[The developer claims that $\sigma = 6$ .
Since $6$ is within our confidence interval of $4.79 < \sigma < 8.43$ , the confidence interval does not contradict this claim.

STEP 9

[The 80% confidence interval for the population standard deviation is $4.79 < \sigma < 8.43$ .
This does not contradict the developer's claim that $\sigma = 6$ .

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Studdy Solution

STEP 1

STEP 2

1. Calculate Degrees of Freedom
2. Find Critical Chi-Squared Values
3. Calculate the Lower Bound of the Confidence Interval
4. Calculate the Upper Bound of the Confidence Interval
5. Check the Claim

STEP 3

We start by figuring out the degrees of freedom, which is related to the sample size.
Since our sample size is $\text{n} = 12$ , the degrees of freedom is $\text{n}-1 = 12 - 1 = \bold{11}$ .
Remember, degrees of freedom basically tells us how much information we're getting from our sample!

STEP 4

For an 80% confidence interval, we have $\alpha = 1 - 0.80 = 0.20$ , so $\alpha/2 = 0.20/2 = \bold{0.10}$ .
We need to find two chi-squared values: $\chi^2_{1-\alpha/2}$ and $\chi^2_{\alpha/2}$ .

STEP 5

With 11 degrees of freedom and $\alpha/2 = 0.10$ , we look up these values in a chi-squared table or use a calculator.
We find $\chi^2_{1-0.10} = \chi^2_{0.90} = \bold{5.578}$ and $\chi^2_{0.10} = \bold{17.275}$ .
These values mark the boundaries of the middle 80% of the chi-squared distribution.

STEP 6

The formula for the lower bound is $\sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{\alpha/2}}}$ .
Plugging in our values, we get [\sqrt{\frac{(12-1) \cdot 6^2}{17.275}} = \sqrt{\frac{11 \cdot 36}{17.275}} = \sqrt{\frac{396}{17.275}} \approx \sqrt{22.91} \approx \bold{4.79}\).

STEP 7

The formula for the upper bound is \[\sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{1-\alpha/2}}}.
Plugging in our values, we get [\sqrt{\frac{(12-1) \cdot 6^2}{5.578}} = \sqrt{\frac{11 \cdot 36}{5.578}} = \sqrt{\frac{396}{5.578}} \approx \sqrt{71.00} \approx \bold{8.43}\).

STEP 8

[The developer claims that $\sigma = 6$ .
Since $6$ is within our confidence interval of $4.79 < \sigma < 8.43$ , the confidence interval does not contradict this claim.

STEP 9

[The 80% confidence interval for the population standard deviation is $4.79 < \sigma < 8.43$ .
This does not contradict the developer's claim that $\sigma = 6$ .

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Studdy Solution

STEP 1

STEP 2

1. Calculate Degrees of Freedom2. Find Critical Chi-Squared Values3. Calculate the Lower Bound of the Confidence Interval4. Calculate the Upper Bound of the Confidence Interval5. Check the Claim

STEP 3

STEP 4

For an 80% confidence interval, we have α=1−0.80=0.20\alpha = 1 - 0.80 = 0.20α=1−0.80=0.20, so α/2=0.20/2=0.10\alpha/2 = 0.20/2 = \bold{0.10}α/2=0.20/2=0.10.We need to find two chi-squared values: χ1−α/22\chi^2_{1-\alpha/2}χ1−α/22​ and χα/22\chi^2_{\alpha/2}χα/22​.

STEP 5

STEP 6

STEP 7

The formula for the upper bound is \[\sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{1-\alpha/2}}}.Plugging in our values, we get [\sqrt{\frac{(12-1) \cdot 6^2}{5.578}} = \sqrt{\frac{11 \cdot 36}{5.578}} = \sqrt{\frac{396}{5.578}} \approx \sqrt{71.00} \approx \bold{8.43}\).

STEP 8

[The developer claims that σ=6\sigma = 6σ=6.Since 666 is within our confidence interval of 4.79<σ<8.434.79 < \sigma < 8.434.79<σ<8.43, the confidence interval *does not* contradict this claim.

STEP 9

[The 80% confidence interval for the population standard deviation is 4.79<σ<8.434.79 < \sigma < 8.434.79<σ<8.43.This *does not* contradict the developer's claim that σ=6\sigma = 6σ=6.

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1. Calculate Degrees of Freedom
2. Find Critical Chi-Squared Values
3. Calculate the Lower Bound of the Confidence Interval
4. Calculate the Upper Bound of the Confidence Interval
5. Check the Claim

For an 80% confidence interval, we have $\alpha = 1 - 0.80 = 0.20$ , so $\alpha/2 = 0.20/2 = \bold{0.10}$ .
We need to find two chi-squared values: $\chi^2_{1-\alpha/2}$ and $\chi^2_{\alpha/2}$ .

The formula for the upper bound is \[\sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{1-\alpha/2}}}.
Plugging in our values, we get [\sqrt{\frac{(12-1) \cdot 6^2}{5.578}} = \sqrt{\frac{11 \cdot 36}{5.578}} = \sqrt{\frac{396}{5.578}} \approx \sqrt{71.00} \approx \bold{8.43}\).

[The developer claims that $\sigma = 6$ .
Since $6$ is within our confidence interval of $4.79 < \sigma < 8.43$ , the confidence interval does not contradict this claim.

[The 80% confidence interval for the population standard deviation is $4.79 < \sigma < 8.43$ .
This does not contradict the developer's claim that $\sigma = 6$ .