Question

We use the t distribution to construct a confidence interval for the population mean when the underlying population standard deviation is not known. Under the assumption that the population is normally distributed, find t_α/2,df for the following scenarios. (You may find it useful to reference the t table. Round your answers to 3 decimal places.)

		tα/2,df
a.	A 95% confidence level and a sample of 10 observations.
b.	A 99% confidence level and a sample of 10 observations.
c.	A 95% confidence level and a sample of 25 observations.
d.	A 99% confidence level and a sample of 25 observations.

Answer 1

solution

a)

n=10

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

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t _/2,df = t_0.025,9 = 2.262 ( using student t table)

b}

n = 10

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

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005,9 = 3.250 ( using student t table

c}

n = Degrees of freedom = df = n - 1 =25 - 1 = 24

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t _/2,df = t_0.025,24 = 2.064 ( using student t table)

d}

n = 25

Degrees of freedom = df = n - 1 = 25- 1 = 24

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005,24 = 2.797 ( using student t table)

		tα/2,df
a.	A 95% confidence level and a sample of 10 observations.	2.262
b.	A 99% confidence level and a sample of 10 observations.	3.250
c.	A 95% confidence level and a sample of 25 observations.	2.064
d.	A 99% confidence level and a sample of 25 observations.	2.797

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