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MARKAKINGSHAWTYIN
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Raiza wants to know the mean of all entering trainees in a boot camp. The mean age of a random sample of 25 trainees is 18 years and the standard deviation is 1.3 years. The sample comes from a normally distributed population. Use a 99% confidence level and find the following:

a. Point estimate
b. Interval Estimate of the population mean

pahelp po with solutions sana. thank you~​

Sagot :

SANTI2922IN

a. 18

b. (17 . 27, 18 . 73)

Step-by-step explanation:

[tex]n = 25 \: \: \: \: x = 18 \: \: \: s = 1.3 \\ \alpha = 1 - 99\% = 1\% \: \: \frac{ \alpha }{2} = 0.5\% \\ t \frac{ \alpha }{2} (n - 1) \: t \: 0.5\%(24) = 2.7969[/tex]

a. Point estimate = x = 18

b.

[tex]x + t \frac{ \alpha }{2} (n - 1). \: s \sqrt[n]{n} \\ = 18 + 2.7969 \times 1.3 \div \sqrt[n]{25} \\ = 18 + 2.7969 \times 1.3 \div 5 \\ \: \: \: \: \: \: \: \: \: \: \: = 18 + 0.727194 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = (17.27 \: \: 18.73)[/tex]

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