Question

Calculate the margin of error and construct the confidence interval for the population mean using the Student's t-distribution (you may assume the population data is normally distributed).

T-Distribution Table

a.  x̄ =87.3, n=64, s=19.6, x̄ =87.3, n=64, s=19.6, 98% confidence

E=E=

Round to two decimal places

  <  μ  <  <  μ  <  

Round to two decimal places

b.  x̄ =31.6, n=44, s=14.6, x̄ =31.6, n=44, s=14.6, 90% confidence

E=E=

Round to two decimal places

  <  μ  <  <  μ  <  

Round to two decimal places

Please provide correct answers. thanks

Answer 1

Solution :

a) degrees of freedom = n - 1 = 64 - 1 = 63

t/2,df = t0.01,63 = 2.387

Margin of error = E = t/2,df * (s /n)

= 2.387 * ( 19.6 / 64)

Margin of error = E = 5.85

The 98% confidence interval estimate of the population mean is,

- E < < + E

87.3 - 5.85 < < 87.3 + 5.85

( 81.45 < < 93.15 )

b) degrees of freedom = n - 1 = 44 - 1 = 43

t/2,df = t0.05,43 = 1.681

Margin of error = E = t/2,df * (s /n)

= 1.681 * ( 14.6 / 44)

Margin of error = E = 3.70

The 90% confidence interval estimate of the population mean is,

- E < < + E

31.6 - 3.70 < < 31.6 + 3.70

( 27.90 < < 35.30 )