baby weights: following are weights, in pounds, of 11 two-month-old baby girls. it is reasonable to assume that the population is approximately normal. use the ti-84 plus calculator as needed.

12.23 12.32 11.87 12.34 11.48 12.66
8.51 12.95 10.30 9.34 8.63

part 1 of 2
(a) construct a 99.9% confidence interval for the mean weight of two-month-old baby girls. round the answers to at least three decimal places.

a 99.9% confidence interval for the mean weight of two-month-old baby girls is $ \square

Question

baby weights: following are weights, in pounds, of 11 two-month-old baby girls. it is reasonable to assume that the population is approximately normal. use the ti-84 plus calculator as needed.

12.23 12.32 11.87 12.34 11.48 12.66
8.51 12.95 10.30 9.34 8.63

part 1 of 2
(a) construct a 99.9% confidence interval for the mean weight of two-month-old baby girls. round the answers to at least three decimal places.

a 99.9% confidence interval for the mean weight of two-month-old baby girls is $ \square < \mu < \square $

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Confidence Interval for Mean", "t-Distribution Critical Value" ], "new_concepts": [], "current_concepts": [ "Confidence Interval for Mean", "t-Distribution Critical Value", "Sample Mean", "Sample Standard Deviation" ] } ## Calculate sample statistics Using the *Confidence Interval for Mean* knowledge point $$ \begin{aligned} n &= 11 \ \sum x &= 12.23 + 12.32 + 11.87 + 12.34 + 11.48 + 12.66 + 8.51 + 12.95 + 10.30 + 9.34 + 8.63 = 122.63 \ \bar{x} &= \frac{122.63}{11} \approx 11.14818 \ \sum x^2 &= 12.23^2 + 12.32^2 + 11.87^2 + 12.34^2 + 11.48^2 + 12.66^2 + 8.51^2 + 12.95^2 + 10.30^2 + 9.34^2 + 8.63^2 \ &= 149.5729 + 151.7824 + 140.8969 + 152.2756 + 131.7904 + 160.2756 + 72.4201 + 167.7025 + 106.0900 + 87.2356 + 74.4769 \ &= 1394.5189 \ s &= \sqrt{\frac{\sum x^2 - \frac{(\sum x)^2}{n}}{n-1}} = \sqrt{\frac{1394.5189 - \frac{122.63^2}{11}}{10}} \approx 1.69176 \end{aligned} $$ ## Find critical value Using the *t-Distribution Critical Value* knowledge point $$ \begin{aligned} df &= n - 1 = 10 \ c &= 0.999 \ \alpha &= 1 - 0.999 = 0.001 \ t_{\alpha/2, df} &= t_{0.0005, 10} \approx 4.587 \end{aligned} $$ ## Compute margin of error Using the *Confidence Interval for Mean* knowledge point $$ \begin{aligned} E &= t_{\alpha/2, df} \cdot \frac{s}{\sqrt{n}} \ E &\approx 4.587 \cdot \frac{1.69176}{\sqrt{11}} \approx 4.587 \cdot 0.51008 \approx 2.3397 \end{aligned} $$ ## Determine confidence interval Using the *Confidence Interval for Mean* knowledge point $$ \begin{aligned} \text{Lower Limit} &= \bar{x} - E \approx 11.14818 - 2.3397 \approx 8.808 \ \text{Upper Limit} &= \bar{x} + E \approx 11.14818 + 2.3397 \approx 13.488 \end{aligned} $$ A 99.9% confidence interval for the mean weight of two-month-old baby girls is $8.808$ $< \mu <$ $13.488$. { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Statistics", "Confidence Interval for Mean" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate sample statistics

Find critical value

Compute margin of error

Determine confidence interval

Answer: