Question

You have ordered 10 bags of cement, which are supposed to weigh 94 kg each. The e eight of the 10 bags is 93.5 kg. Assuming thad the 10 weights can be viewed as a realization of a random sample from a normal distribution with unknown parameters, construct a 95% confidence interval for the expected weight of a bag. The sample standard deviation of the 10 weights is 0.75. еат

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 93.5

sample standard deviation = s = 0.75

sample size = n = 10

Degrees of freedom = df = n - 1 = 10 - 1 = 9

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_0.025,9 = 2.262

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.262 * (0.75 / $\sqrt$10)

= 0.536

The 95% confidence interval is,

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

93.5 - 0.536 < $\mu$ < 93.5 + 0.536

92.964 < $\mu$ < 94.036

(92.964 , 94.036)