Question

Say that a sample of size 49 taken from a normally distributed population, has variance = 1. The sample has a mean of 0.49. Then, a 95% confidence interval for µ will be

		0.49 – (1.96)(1/7) < µ < 0.49 + (1.96)(1/7)
		0.49 – (2.575)(1/7) < µ < 0.49 + (1.96)(1/7)
		49 – (1.96)(1/0.7) < µ < 49 + (1.96)(1/0.7)
		49 – (2.575)(1/0.7) < µ < 49 + (2.575)(1/0.7)

Answer 1

Solution :

Given that,

= 0.49

² = 1

= 1

n = 49

At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* (/n)

= 1.96 * (1 / 49 )

= 0.28

At 95% confidence interval estimate of the population mean is,

- E < < + E

0.49 - 0.28 < < 0.49 + 0.28

0.21 < < 0.77

Option 0.49 - ( 1.96 ) ( 1/ 7) < µ < 0.49 + (1.96)(1/7) is correct.