Question

Calcium is essential to tree growth because it promotes the formation of wood and maintains cell walls. In​ 1990, the concentration of calcium in precipitation in a certain area was 0.11 milligrams per liter​ (mg/L). A random sample of 10 precipitation dates in 2007 results in the following data table. Complete parts​ (a) through​ (c) below.

0.073
0.082
0.086
0.231
0.129
0.187
0.133
0.207
0.322
0.092

Construct a 95% confidence interval about the sample mean concentration of calcium precipitation.

The lower bound is

The upper bound is

Answer 1

Solution

Given that,

x	x2
0.073	0.0053
0.082	0.0067
0.86	0.0074
0.231	0.0534
0.129	0.0166
0.187	0.035
0.133	0.0177
0.207	0.0428
0.322	0.1037
0.092	0.0085
$\sum$x = 1.542	$\sum$x2 = 0.2971

a ) The sample  mean is $\bar x$

Mean is $\bar x$ = $\sum$x / n

= 0.073+0.082+0.086+0.231+0.129+0.187+0.133+0.207+0.322+0.092 / 10

=1.542 / 10

=0.1542

Mean is $\bar x$ = 0.1542

b ) The sample standard is S

S = $\sqrt{}$ ( $\sum$ x² ) - (( $\sum$x )² / n ) / 1 -n )

   = $\sqrt{}$ (0.2971 ( (0.1542 )² / 10) / 9

   = $\sqrt{}$ ( 0.2971 -0.2378 ) / 9

=$\sqrt{}$ (0.0593/ 9 )

= $\sqrt{}$ 0.0066

= 0.0812

The sample standard is = 0.0812

$\bar x$ = 0.1542

s = 0.0812

n = 10

Degrees of freedom = df = n - 1 = 10- 1 = 9

At 95% confidence level the t is,

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

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,10 =2.262

The critical value = 2.262

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.262. * ( 0.0812 / $\sqrt$10 )

= 0.058

Margin of error = 0.058

The 95% confidence interval estimate of the population mean is,

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

0.154 - 0.058 < $\mu$ < 0.154 + 0.058

0.096 < $\mu$ < 0.212

The lower bound is  0.096

The upper bound is  0.212