Question

7 mark QUESTION 24. The breaking strength of a random sample of 20 bundles of wool fibres have a sample mean of 416.5 (Mpa) and standard deviation 12.9 (Mpa). Construct 90% two-sided confidence interval for the breaking strength population mean in Mpa. Assume that the population is normally distributed. a) [415.38, 417.62] b) [411.7, 421.25] c) [411.51, 421.49] d) [00 421.25]: e) None of the above.

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 416.5

Population standard deviation = $\sigma$ = 12.9

Sample size = n = 20

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90$\alpha$ = 0.10

$\alpha$ / 2 = 0.10 / 2 = 0.05

Z_$\alpha$/2 = Z_0.05 = 1.645

Margin of error = E = Z_$\alpha$/2* ($\sigma$ /$\sqrt$n)

=1.645 * (12.9 / $\sqrt$ 20)

= 4.8

At 99% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

416.5 - 4.8 < $\mu$ < 416.5 + 4.8

411.7 < $\mu$ < 421.3

( 411.7 , 421.3 )

option b) is correct