Question

answers and explanations

A botanist measures the heights of 25 seedlings and obtains a mean of 41.6 cm and standard deviation of 4.8 cm. 9. What is the best point estimate for the population mean? 10. Construct a 98% confidence interval for the population mean. 11. Construct a 95% confidence interval for the population standard deviation.

Answer 1

Solution:

Given ,

n = 25

$\bar x$ = 41.6

s = 4.8

9)

The sample mean is the best point estimate of the population mean.

So , required best point estimate is 41.6

10)

Our aim is to construct 98% confidence interval.   

$\therefore$ c = 0.98

$\therefore$$\alpha$ = 1- c = 1- 0.98 = 0.02

$\therefore$  $\alpha$/2 = 0.10 $\slash$ 2 = 0.01

Also, d.f = n - 1 = 25 - 1 = 24

$\therefore$  
ta/2.0.f.
  =  
ta/2,1-1
  =  _{0.01 , 24} =  2.492

( use t table or t calculator to find this value..)

The margin of error is given by

E =  $\alpha$/2,d.f. * ( / $\sqrt{}$ n)

=  2.492 * (4.8 / $\sqrt{}$ 25)

= 2.39232

Now , confidence interval for mean($\mu$) is given by:

($\bar x$ - E ) <  $\mu$ <  ($\bar x$ + E)

(41.6 - 2.39232)   <  $\mu$ <  (41.6 + 2.39232)

39.20768 <  $\mu$ <  43.99232

Required 98% confidence interval for mean is (39.20768 , 43.99232)

11)

df = n - 1 = 24

At 95% confidence level ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.025 and 1 - ($\alpha$ / 2) = 0.975

Now , using chi square table ,

$\chi$²_{$\alpha$/2,df} = 39.36

$\chi$²_{1 - $\alpha$ /2,df} = 12.40

The 95% confidence interval for $\sigma$ is,

$s\sqrt{\frac{n-1}{\chi_{\alpha/2,df}^{2}}} <\sigma< s\sqrt{\frac{n-1}{\chi_{1 - \alpha/2,df}^{2}}}$

4.8$\sqrt{}$[( 25 - 1 ) / 39.36] < $\sigma$ < 4.8$\sqrt{}$[( 25 - 1 ) / 12.40]

3.75 < $\sigma$ < 6.68