Math  /  Data & Statistics

QuestionYou work for a soft-drink company in the quality control division. You are interested in the standard deviation of one of your production lines as a measure of consistency. The product is intended to have a mean of 12 ounces, and your team would like the standard deviation to be as low as possible. You gather a random sample of 18 containers. Estimate the population standard deviation at a 98%98 \% level of confidence. \begin{tabular}{|r|r|r|r|r|r|} \hline 11.99 & 12.13 & 11.86 & 12.01 & 11.95 & 12.11 \\ \hline 11.88 & 11.91 & 12 & 11.98 & 12.09 & 12.13 \\ \hline 11.98 & 11.93 & 12.14 & 12.08 & 11.98 & 12.07 \\ \hline \end{tabular} (Data checksum: 216.22) Note: Keep as many decimals as possible while making these calculations. If possible, keep all answers exact by storing answers as variables on your calculator or computer. a) Find the sample standard deviation: 0.0836×0.0836 \times \square b) Find the lower and upper χ2\chi^{2} critical values at 98%98 \% confidence:
Lower: 6.4077 No66.4077 \mathrm{~N} \mathrm{o}^{6} Upper: 30.1911 x c) Report your confidence interval for σ:(0.0569×,0.1152×)\sigma:(0.0569 \times, 0.1152 \times)

Studdy Solution

STEP 1

What is this asking? We need to estimate how much the drink amounts vary in our production line, using a sample of 18 drinks, and we want to be 98% sure our estimate is correct. Watch out! Don't mix up standard deviation with variance (standard deviation squared).
Also, remember the Chi-Squared distribution is not symmetrical!

STEP 2

1. Calculate the sample standard deviation.
2. Find the critical Chi-Squared values.
3. Calculate the confidence interval for the population standard deviation.

STEP 3

First, let's **calculate the sample mean**, xˉ\bar{x}.
Add up all the drink amounts and divide by the number of drinks, which is **18**. 11.99+12.13+11.86++12.07=216.2211.99 + 12.13 + 11.86 + \dots + 12.07 = 216.22.
Dividing by **18** gives us xˉ=216.2218=12.0122\bar{x} = \frac{216.22}{18} = 12.0122.
Remember this **mean**!

STEP 4

Now, **calculate the sample variance**, s2s^2.
For each drink amount, subtract the **mean** we just calculated, square the result, and add all those squared differences together.
Then, divide by n1=181=17n-1 = 18-1 = 17.
This gives us s2=(11.9912.0122)2+(12.1312.0122)2++(12.0712.0122)217=0.119817=0.0070s^2 = \frac{(11.99 - 12.0122)^2 + (12.13 - 12.0122)^2 + \dots + (12.07 - 12.0122)^2}{17} = \frac{0.1198}{17} = 0.0070.

STEP 5

Finally, **take the square root of the sample variance** to get the **sample standard deviation**, ss.
So, s=0.0070=0.0837s = \sqrt{0.0070} = 0.0837.
This tells us how spread out our sample data is!

STEP 6

Since we want a **98% confidence interval**, we have 10.98=0.021 - 0.98 = 0.02 left over for the tails.
We split this equally between the two tails, getting 0.022=0.01\frac{0.02}{2} = 0.01 in each tail.

STEP 7

We have n1=181=17n - 1 = 18 - 1 = 17 degrees of freedom.
Using a Chi-Squared table or calculator, we find the **critical values** for 0.010.01 and 0.990.99 (1 minus the left tail) with **17 degrees of freedom**.
The **lower critical value** is 6.4086.408 and the **upper critical value** is 33.40933.409.

STEP 8

The formula for the confidence interval is (n1)s2χupper2<σ<(n1)s2χlower2\sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{\text{upper}}}} < \sigma < \sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{\text{lower}}}}.
We plug in our values: n=18n = 18, s2=0.0070s^2 = 0.0070, χlower2=6.408\chi^2_{\text{lower}} = 6.408, and χupper2=33.409\chi^2_{\text{upper}} = 33.409.

STEP 9

**Calculate the lower bound:** (181)0.007033.409=0.11933.409=0.0036=0.0600\sqrt{\frac{(18-1) \cdot 0.0070}{33.409}} = \sqrt{\frac{0.119}{33.409}} = \sqrt{0.0036} = 0.0600.

STEP 10

**Calculate the upper bound:** (181)0.00706.408=0.1196.408=0.0186=0.1364\sqrt{\frac{(18-1) \cdot 0.0070}{6.408}} = \sqrt{\frac{0.119}{6.408}} = \sqrt{0.0186} = 0.1364.

STEP 11

So, we are 98% confident that the true population standard deviation of the drink amounts is between **0.0600 ounces** and **0.1364 ounces**.

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