Question

A car manufacturer is studying the rockwell hardness of its seat recliner mechanism. A sample of 35 cars has an average of 41 .50 and standard deviation of 0.420. a) What is the 90 % confidence interval to estimate the true rockwell hardness? b) What is the 95% confidence interval to estimate the true rockwell hardness?

Answer 1

Solution :

Given that,

a)

sample size = n = 35

Degrees of freedom = df = n - 1 = 35 - 1 = 34

t_{$\alpha$ /2,df} = 1.691

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

= 1.691 * (0.420 / $\sqrt$35)

= 0.12

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

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

41.50 - 0.12 < $\mu$ < 41.50 + 0.12

41.38 < $\mu$ < 41.62

The 90% confidence interval to estimate the true Rockwell hardness: 41.38 , 41.62

b)

sample size = n = 35

Degrees of freedom = df = n - 1 = 35 - 1 = 34

t_{$\alpha$ /2,df} = 2.032

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

= 2.032 * (0.420 / $\sqrt$35)

= 0.14

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

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

41.50 - 0.14 < $\mu$ < 41.50 + 0.14

41.36 < $\mu$ < 41.64

The 95% confidence interval to estimate the true Rockwell hardness: 41.36 , 41.64