Question 21 [t-interval] a random sample of size

Ответ:

hilllz4793

14.03.2022

Option a. [52.78,64.22]

Step-by-step explanation:

We are given that a random sample of size 18 is drawn from a population that is normally distributed.

Sample mean is 58.5, and the sample standard deviation is found to be 11.5 i.e., X bar = 58.5 and s = 11.5

The pivotal quantity for calculating 95% confidence interval is;

$\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }$ ~ t_n_-_1

So, 95% confidence interval about population mean is given by;

P(-2.110 < t_1_7 < 2.110) = 0.95

P(-2.110 < $\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }$ < 2.110) = 0.95

P(-2.110 * $\frac{s}{\sqrt{n} }$ < ${Xbar - \mu}$ < 2.110 * $\frac{s}{\sqrt{n} }$ ) = 0.95

P(X bar - 2.110 * $\frac{s}{\sqrt{n} }$ < $\mu$ < X bar + 2.110 * $\frac{s}{\sqrt{n} }$ ) = 0.95

95% confidence interval about $\mu$ = [ X bar - 2.110 * $\frac{s}{\sqrt{n} }$ , X bar + 2.110 * $\frac{s}{\sqrt{n} }$ ]

= [ 58.5 - 2.110 * $\frac{11.5}{\sqrt{18} }$ , 58.5 + 2.110 * $\frac{11.5}{\sqrt{18} }$ ]

= [ 52.78 , 64.22 ]

Therefore, 95% confidence interval about population mean is [52.78 , 64.22].

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