Mathematics, 08.11.2019 05:31 sg3635939

Agroup of engineers developed a new design for a steel cable. they need to estimate the amount of weight the cable can hold. the weight limit will be reported on cable packaging. the engineers take a random sample of 41 cables and apply weights to each of them until they break. the 41 cables have a mean breaking weight of 772.3 lb. the standard deviation of the breaking weight for the sample is 15.3 lb. find the 95% confidence interval to estimate the mean breaking weight for this type cable. ( 11.74 , 12.86 ) your answer should be to 2 decimal places.

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Answer from: madleneinejessup

The confidence interval is a one of the estimates that can be determined from an observed data

The confidence interval is: $\mathbf{ (767.47,777.13)}$

The given parameters are:

$\mathbf{ n = 41}$

$\mathbf{CI = 95\%}$

$\mathbf{\mu = 772.3}$

$\mathbf{\sigma = 15.3}$

Start by calculating the standard deviation of the sample

$\mathbf{\sigma_x = \frac{\sigma}{\sqrt n}}$

$\mathbf{\sigma_x = \frac{15.3}{\sqrt{41}}}$

$\mathbf{\sigma_x = \frac{15.3}{6.40}}$

$\mathbf{\sigma_x = 2.39}$

Next, calculate the degree of freedom (df)

$\mathbf{ df = n-1}$

So, we have:

$\mathbf{ df = 41-1}$

$\mathbf{ df = 40}$

Calculate the significance level

$\mathbf{\alpha = \frac{1 - CI}{2}}$

$\mathbf{\alpha = \frac{1 - 95\%}{2}}$

$\mathbf{\alpha = \frac{0.05}{2}}$

$\mathbf{\alpha = 0.025}\\$

The T value, when $\mathbf{\alpha = 0.025}\\$ and $\mathbf{ df = 40}$ us:

$\mathbf{T = 2.021}$

The endpoints of the confidence interval is calculated as:

$\mathbf{CI = \mu \pm T \times \sigma}$

So, we have:

$\mathbf{CI = 772.3 \pm 2.021 \times 2.39}$

$\mathbf{CI = 772.3 \pm 4.83}$

Split

$\mathbf{CI = (772.3 - 4.83,772.3 + 4.83)}$

$\mathbf{CI = (767.47,777.13)}$

Hence, the confidence interval is: $\mathbf{ (767.47,777.13)}$

Read more about confidence intervals at:

link

Answer from: Isabella1319

The 95% confidence interval for the mean breaking weight for this type cable is (767.47 lb, 777.13 lb).

Step-by-step explanation:

Our sample size is 41

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

df = 41-1 = 40

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 40 and 0.025 in the two-sided t-distribution table, we have T = 2.021

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{15.3}{\sqrt{41}} = 2.39$

Now, we multiply T and s

M = T*s =2.021*2.39 = 4.83

Then

The lower end of the confidence interval is the mean subtracted by M. So:

L = 772.3 - 4.83 = 767.47

The upper end of the confidence interval is the mean added to M. So:

L = 772.3 + 4.83 = 777.13

The 95% confidence interval for the mean breaking weight for this type cable is (767.47 lb, 777.13 lb).

Answer from: Quest

answer: there are 453.592 grams in a pound.

Answer from: Quest

a. turk and mcalister

b. howard, mission, and market

c. yes. since streets end some do not cross yet are not parallel to others. example: mcalister and 5th

step-by-step explanation:

parallel lines are lines which do not intersect ever and look a lot like railroad tracks. turk, mcalister and golden gate are examples of these.

intersect are streets which cross each other and a car on one can reach the other. howard, mission, and market intersect with 6th.

skew lines are lines which neither intersect nor are parallel. these lines exist here because some streets end and never intersect another like mcalister and 5th.

A. name the streets that are parallel to golden gate. b. name the streets that intersect 6th street.

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Agroup of engineers developed a new design for a steel cable. they need to estimate the amount of we...

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