Question

The owner of the House of Greens Greenhouse wants to estimate the mean height that her seedlings grow. A sample of 29 plants were chosen and their growth was recorded over a period of a month. It was found that the sample mean was 20.00 cm and the sample standard deviation was 2.25 cm. Given this information, develop a 99.0% confidence interval estimate for the population mean (assuming the best point estimate for the population mean is the sample mean).

Answer 1

solution

Given that,

= 20.00

s =2.25

n = 29

Degrees of freedom = df = n - 1 =29 - 1 = 28

At 99.0% confidence level the t is ,

= 1 - 99.0% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,28 = 2.763    ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.763 * (2.25 / 29) = 1.1544

The 99.0% confidence interval estimate of the population mean is,

- E < < + E

20.00 -1.1544 < < 20.00+ 1.1544

18.8456 < < 21.1544

( 18.8456 ,21.1544