Question

QUESTION 1 1 points Saved Copy of A research engineer for a tire manufacturer is investigating tire life for a new rubber compound and has built 36 tires and tested them to end-of life in a road test. The sample mean and standard deviation are 66 and 2.1 thousand kilometers. Find a 99% upper bound for the mean life in road test. (round the answer to three digits) 67.853 QUESTION 2 1 points Saved A synthetic fiber used in manufacturing carpet has tensile strength that is normally distributed with mean 75 psi and standard deviation 3 psi. Find the probability that a random sample of n 16 fiber specimens will have sample mean tensile strength that exceeds 76 psi. 0.0918

Answer 1

Solution :

Given that,

Q-1

Point estimate = sample mean = $\bar x$ = 66.9

sample standard deviation = s = 2.1

sample size = n = 36

Degrees of freedom = df = n - 1 = 35

At 99% confidence level the t is ,

$\alpha$ = 1 - 99% = 1 - 0.95 = 0.01

$\alpha$ / 2 = 0.01/ 2 = 0.005

t_{$\alpha$ /2,df} = t_0.005,35 = 2.724

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

= 2.724 * (2.1 / $\sqrt$36)

= 0.953

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

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

66.9 - 0.953 < $\mu$ < 66.9 + 0.953

65.947 < $\mu$ < 67.853

Upper bound for the mean life is 67.853

Q-2

mean = $\mu$ = 75

standard deviation = $\sigma$ = 3

n = 16

$\mu$_{$\bar x$} = $\mu$ = 75 and

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 3 / $\sqrt$ 16 = 0.75

P($\bar x$ > 76) = 1 - P($\bar x$ < 76)

= 1 - P(($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (76-75) / 0.75)

= 1 - P(z < 1.33)

= 1 - 0.9082

= 0.0918

Probability = 0.0918

Answer 2

Solution :

Given that,

Q-1

Point estimate = sample mean = $\bar x$ = 66.9

sample standard deviation = s = 2.1

sample size = n = 36

Degrees of freedom = df = n - 1 = 35

At 99% confidence level the t is ,

$\alpha$ = 1 - 99% = 1 - 0.95 = 0.01

$\alpha$ / 2 = 0.01/ 2 = 0.005

t_{$\alpha$ /2,df} = t_0.005,35 = 2.724

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

= 2.724 * (2.1 / $\sqrt$36)

= 0.953

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

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

66.9 - 0.953 < $\mu$ < 66.9 + 0.953

65.947 < $\mu$ < 67.853

Upper bound for the mean life is 67.853

Q-2

mean = $\mu$ = 75

standard deviation = $\sigma$ = 3

n = 16

$\mu$_{$\bar x$} = $\mu$ = 75 and

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 3 / $\sqrt$ 16 = 0.75

P($\bar x$ > 76) = 1 - P($\bar x$ < 76)

= 1 - P(($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (76-75) / 0.75)

= 1 - P(z < 1.33)

= 1 - 0.9082

= 0.0918

Probability = 0.0918