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3 years ago

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A researcher is interested in estimating the mean weight of a semi trailer truck to determine the potential load capacity. She t

akes a random sample of 17 trucks and computes a sample mean of 20,000 pounds with sample standard deviation of 1,500. The 95% confidence interval for the population mean weight of a semi trailer truck is ______________.

1 answer:

Kaylis [27]3 years ago

3 0

Answer: 95% confidence interval = 20,000 ± 2.12 $\times$ $\frac{1500}{\sqrt{17} }$

( 19228.736 , 20771.263 ) OR ( 19229 , 20771 )

Step-by-step explanation:

Given :

Sample size(n) = 17

Sample mean = 20000

Sample standard deviation = 1,500

5% confidence

∴ $\frac{\alpha}{2}$ = 0.025

Degree of freedom ( $d_{f}$ ) = n-1 = 16

∵ Critical value at ( 0.025 , 16 ) = 2.12

∴ 95% confidence interval = mean ± $Z_{c}$ $\times$ $\frac{\sigma}{\sqrt{n} }$

Critical value at 95% confidence interval = 20,000 ± 2.12 $\times$ $\frac{1500}{\sqrt{17} }$

( 19228.736 , 20771.263 ) OR ( 19229 , 20771 )

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3 years ago

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KATRIN_1 [288]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e ( $x_o, y_o, z_o$ ) and are parallel to $\dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}$

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

$x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct$

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

$\begin {vmatrix} x = 4+ t \\ \\ \dfrac{x-4}{1} = t \begin {vmatirx} \end {vmatrix}$ $\begin {vmatrix} y = - 5+2t \\ \\ \dfrac{y+5}{2} =t \end {vmatrix}$ $\begin {vmatrix} z =2+t \\ \\ \dfrac{z-2}{1} =t \end {vmatrix}$

∴

$\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}$

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3 years ago

A random sample of 5120 permanent dwellings on an entire reservation showed that 1627 were traditional hogans.

krek1111 [17]

Answer:

a) $\hat p=\frac{1627}{5120}=0.3178$

b)The 99% confidence interval would be given by (0.301;0.355)

Step-by-step explanation:

1) Notation and definitions

$X=1627$ number of permanent dwellings on the entire reservation that are traditional hogans.

$n=5120$ random sample taken

$\hat p=\frac{1627}{5120}=0.318$ estimated proportion of permanent dwellings on the entire reservation that are traditional hogans.

$p$ true population proportion of permanent dwellings on the entire reservation that are traditional hogans.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Part a

The point of estimate for p=the proportion of all permanent dwellings on the entire reservation that are traditional hogans is given by:

$\hat p=\frac{1627}{5120}=0.3178$

Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.318 - 2.58\sqrt{\frac{0.318(1-0.318)}{5120}}=0.301$

$0.318 + 2.58\sqrt{\frac{0.318(1-0.318)}{5120}}=0.335$

The 99% confidence interval would be given by (0.301;0.355)

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