Answer to Question #307946 in Statistics and Probability for ASH

Question

A random sample of size 15 has 50 as mean, the sum of the

squares of the deviation taken from mean is 130. Can this

sample be regarded as taken from the population having 53 as

mean? Obtain 95% and 99% confidence limits of the mean

for the population.(t test example)

Answer 1

First, we compute the sample standard deviation where:

$s = \sqrt{\dfrac{\sum_{i=1}^{n}(x_i - \overline{x})^{2}}{n - 1}}$

Given that $SS = \sum_{i=1}^{n}(x_i - \overline{x})^{2}$ = 130

and n=15

then:

s = $\sqrt\frac{130}{14}$ = 3.047

Next, we compute the confidence interval using the formula:

CI = x̅ ± t*$\frac{s}{\sqrt{n}}$

The degrees of freedom = n-1 = (15-1) = 14

For 14 degrees of freedom and 95% confidence interval, we find that the critical value t= ±2.14 (two tailed test)

Therefore, the 95% confidence interval is:

95% CI = 50± 2.14 *$\frac{3.047}{\sqrt{15}}$ = 50± 1.68

The lower limit = 50-1.68 = 48.32

The upper limit = 50+1.68 = 51.68

The 95% confidence interval is (48.32, 51.68)

Also,

For 14 degrees of freedom and 99% confidence interval, we find that the critical value t= ±2.98 (two tailed test)

Therefore, the 99% confidence interval is:

99% CI = 50± 2.98 *$\frac{3.047}{\sqrt{15}}$ = 50± 2.34

The lower limit = 50-2.34 = 47.66

The upper limit = 50+2.34 = 52.34

The 99% confidence interval is (47.66, 52.34)

Since the population mean (53) is not contained within any of the two confidence intervals for the sample mean, we conclude that the sample may not be regarded as being taken from the population having 53 as

mean