Question

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Answer 1

Given, a random sample of 1400 seniors spend an average of $4900 on medicines with standard deviation $900. i.e.

The sample size = 1400

The sample mean = 4900

(1)

The sample standard deviation = 900

(s)

We have to find the population mean with an confidence interval 95%.

(1)

C.I = 0.95 a=1-0.95 = 0.5 = 0.25

**Note, when the population standard deviation$(\sigma)$ is not known and sample size is small, then the mean has a student's t (t) distribution and the sample standard deviation
(s)
is used instead of the population standard deviation $(\sigma)$ .

But here the sample size is large enough, hence the population standard deviation estimate is taken to be same as the sampke standard deviation. i.e. $\hat{\sigma}=s$ .

So the standard error is given by,

$S.E =\frac{\sigma}{\sqrt{n}}= \frac{s}{\sqrt{n}}=\frac{900}{\sqrt{1400}} \approx 24.0535$

Then the estimation for population mean with 95% confidence interval is given by,

$|\mu - \bar{x}| < z_{\alpha/2}\ S.E$

From z-table we get,

$z_{\alpha/2}=z_{0.25}=1.96$

and hence we get,

$\mathrm{\bar{x}- z_{0.25}\ \frac{s}{\sqrt{n} }< \mu < \bar{x}+ z_{0.25}\ \frac{s}{\sqrt{n}}}$

$\mathrm{ \implies 4900- 1.96\times 24.0535< \mu < 4900+ 1.96\times 24.0535}$  

$\implies \mathrm{ 4852.8551< \mu <4947.1448}$

Therefore the estimate is, $4853 to $4917.