Question

If a sample of size n = 25 is chosen from a normal population, which one of the following is a 95% confidence interval for the population mean when the sample mean is 50 and the sample standard deviation is 6.5? O A 50 2.06(1.3) B 50 2.06(6.5) C 50 2.064(65) D 50 t 2.064(1.3)

Answer 1

Solution :

Given that,

= 50

Given that, x^- = 50 s = 6.5 n = 25 Degrees of freedom = df = n - 1 = 25 - 1 = 24 At 95% confidence level the t is, alpha = 1 - 95% = 1 - 0.95 = 0.05 alpha/2 = 0.05/2 = 0.025 t_alpha/2, df = t_0.025,24 = 2.064 standard error = s/squareroot n = 6.5/squareroot 25 = 6.5/5 = 1.3 The 95% confidence interval for mu is, x^- +/- t_alpha/2, df * S.E. 50 +/- 2.064(1.3) Option D) is correct.

s = 6.5

n = 25

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

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t _{$\alpha$ /2,df} = t_0.025,24 = 2.064

standard error = s / $\sqrt$ n = 6.5 / $\sqrt$ 25 = 6.5 / 5 = 1.3

The 95% confidence interval for $\mu$ is ,

$\bar x$ +/- t _{$\alpha$ /2,df *} S.E.

50 +/- 2.064(1.3)

Option D) is correct .