Question

An ecologist randomly samples 16 plants of a specific species and measures their heights. He finds that this sample has a mean of 13 cm and a standard desabon of 1 cm. If we assure that the height measurements are normally distributed, find a 95% confidence interval for the mean height of all plants of this species. Then complete the table below Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas) What is the lower limit of the confidence interven What is the upper limit of the confidence interval X ?

Answer 1

Solution:

Given:

Sample size = n = 16

Sample mean = = 13 cm

T

Sample Standard deviation = s =1 cm

Confidence level = c = 95%

Formula:

$(\bar{x}-E \: \: < \mu < \: \: \bar{x}+E )$

where

$E=t_{c} \times s / \sqrt{n}$

t_c is t critical value for c = 95%  confidence level

Thus two tail area = 1 - c = 1 - 0.95 = 0.05

df = n - 1 =  16 - 1 = 15

Look in  t table for df = 15 and two tail area = 0.05 and find t critical value

t Table cum. prob one-tail two-tails t 50 0.50 1.00 5.76 0.25 0.50 t.80 0.20 0.40 t.86 0.15 0.30 t.ee 0.10 0.20 tas 0.05 0.10 t.976 0.025 0.05 df UA 1 0.000 2 0.000 3 0.000 4 0.000 5 0.000 6 0.000 7 0.000 8 0.000 9 0.000 10 0.000 11 0.000 12 0.000 13 0.000 14 0.000 15 -0.000 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.601 1.376 1.061 0.978 0.941 0.920 0.906 0.896 0.889 0.883 0.879 0.876 0.873 0.870 0.868 0.8.66 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 1.100 1.093 1.088 1.083 1.079 1.076 4.074 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 1.345 1.34T 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.796 1.782 1.771 1.761 11.71 4.B03 3. 182 2.776 2.571 2.147 2.365 2 306 2262 2.228 2.01 2.179 2.60 2.145 2.131

t_c = 2.131

thus

$E=t_{c} \times s / \sqrt{n}$

$E=2.131 \times 1 / \sqrt{16}$

$E=2.131 \times 1 / 4$

$E=0.53275$

$E=0.5$

thus

$(\bar{x}-E \: \: < \mu < \: \: \bar{x}+E )$

$(13-0.5 \: \: < \mu < \: \: 13+0.5 )$

$(12.5 \: \: < \mu < \: \: 13.5 )$

Thus

Lower Limit = 12.5

Upper limit = 13.5