Question

You intend to estimate a population mean mu with the following sample. 45.3, 42.3, 53, 49, 15.2, 52.3, 45.6, 39.6, 39.4, 16.1, 54.4.You believe the population is normally distributed. Find the 99.9% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places (because the sample data are reported accurate to one decimal place).

Answer 1

Answer:

A 99.9% confidence interval for the population mean is [22.31, 59.91] .

Step-by-step explanation:

We are given the following sample values;

X = 45.3, 42.3, 53, 49, 15.2, 52.3, 45.6, 39.6, 39.4, 16.1, 54.4.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                       P.Q.  =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~  $t_n_-_1$

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$  = 41.11

             s = sample standard deviation = $\sqrt{\frac{\sum (X - \bar X)^{2} }{n-1} }$  =  13.59

             n = sample size = 11

Here for constructing a 99.9% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 99.9% confidence interval for the population mean, $\mu$ is;

P(-4.587 < $t_1_0$ < 4.587) = 0.999  {As the critical value of t at 10 degrees of

                                                freedom are -4.587 & 4.587 with P = 0.05%}    

P(-4.587 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 4.587) = 0.999

P( $-4.587 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}{$ < $4.587 \times {\frac{s}{\sqrt{n} } }$ ) = 0.999  

P( $\bar X-4.587 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+4.587 \times {\frac{s}{\sqrt{n} } }$ ) = 0.999

99.9% confidence interval for $\mu$ = [ $\bar X-4.587 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+4.587 \times {\frac{s}{\sqrt{n} } }$ ]

                                       = [ $41.11-4.587 \times {\frac{13.59}{\sqrt{11} } }$ , $41.11+4.587 \times {\frac{13.59}{\sqrt{11} } }$ ]

                                       = [22.31, 59.91]

Therefore, a 99.9% confidence interval for the population mean is [22.31, 59.91] .