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Tasya [4]
3 years ago
13

Use the standard normal distribution or the​ t-distribution to construct a 99​% confidence interval for the population mean. Jus

tify your decision. If neither distribution can be​ used, explain why. Interpret the results. In a random sample of 42 ​people, the mean body mass index​ (BMI) was 28.3 and the standard deviation was 6.09.
Mathematics
1 answer:
vichka [17]3 years ago
3 0

Answer:

(25.732,30.868)

Step-by-step explanation:

Given that in a random sample of 42 ​people, the mean body mass index​ (BMI) was 28.3 and the standard deviation was 6.09.

Since only sample std deviation is known we can use only t distribution

Std error = \frac{s}{\sqrt{n} } =\frac{6.09}{\sqrt{42} } \\=0.9397

df = 42-1 =41

t critical for 99% two tailed = 2.733

Margin of error= 2.733*0.9397=2.568

Confidence interval lower bound = 28.3-2.568=25.732

Upper bound = 28.3+2.568=30.868

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\red { \orange {\boxed {\boxed{Answer}}}}

<h2>=  \frac{5}{6}  \times  \frac{6}{7}</h2><h2>=  \frac{5 \times 6}{6 \times 7}</h2><h2>=  \frac{30}{42}</h2><h2>=  \frac{30 \div 6}{42 \div 6}</h2><h2>=  \frac{5}{7}</h2>

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2 years ago
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5 0
3 years ago
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