Answer to Question #150174 in Statistics and Probability for caarino

Question

a. A statistics practitioner randomly sampled 100 observations from a population with a standard deviation of 5 and found that x is 10. Estimate the population mean with 90% confidence.

b. Repeat part (a) with a sample size of 25.

c. Repeat part (a) with a sample size of 10.

d. Describe what happens to the confidence interval estimate when the sample size decreases.

Answer 1

a. n = 100

σ = 5

$\bar{x}=10$

For the confidence level 1 – α = 0.90. determine $t_{α/2} = t_{0.05}$ using t table, which is given in the row with df = n-1 = 100-1 = 99 and in the column with $t_{0.05}$ :

$t_{α/2} = 1.64$

The margin of error is then:

$E = t_{α/2} \times \frac{σ}{\sqrt{n}} \\ E = 1.64 \times \frac{5}{\sqrt{100}} = 0.82$

The boundaries of the confident interval:

10 ± 0.82 (9.18 to 10.82)

b. n = 25

For the confidence level 1 – α = 0.90. determine $t_{α/2} = t_{0.05}$ using t table, which is given in the row with df = n-1 = 25-1 = 24 and in the column with $t_{0.05}$ :

$t_{α/2} = 1.71$

The margin of error is then:

$E = t_{α/2} \times \frac{σ}{\sqrt{n}} \\ E = 1.71 \times \frac{5}{\sqrt{25}} = 1.71$

The boundaries of the confident interval:

10 ± 1.71 (8.29 to 11.71)

c. n=10

For the confidence level 1 – α = 0.90. determine $t_{α/2} = t_{0.05}$ using t table, which is given in the row with df = n-1 = 10-1 = 9 and in the column with $t_{0.05}$ :

$t_{α/2} = 1.83$

The margin of error is then:

$E = t_{α/2} \times \frac{σ}{\sqrt{n}} \\ E = 1.83 \times \frac{5}{\sqrt{10}} = 2.89$

The boundaries of the confident interval:

10 ± 2.89 (7.11 to 12.89)

d. Notice how the interval increases and the sample size decreases