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2 years ago

11

If a factory produces 1,672 steering wheels during an 11 hour shift , what is the production rate of steering wheels per hour? [

PLS ANSWER!]

1 answer:

d1i1m1o1n [39]2 years ago

3 0

152 per hour. Just take the 1672 and divide it by 11 to get your rate of change

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Which of the following points are in the fourth quadrant of the xy-plane? Check all that apply.

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The fourth quadrant contains a positive number for the x axis and a negative number for the y axis. (B=4,-5 and F=4,-7)

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3 years ago

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3 years ago

. A study of one thousand teens found that the number of hours they spend on social networking sites each week is normally distr

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We are given the following data:

Sample Size = n = 1000

Sample Mean = u = 15 hours

Population Standard Deviation = s = 2 hours

Since we know the population standard deviation, we can use z distribution to find the confidence interval about the mean. z value for 95% confidence interval is 1.96.

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$(u-z\frac{s}{\sqrt{n}} , u+z\frac{s}{\sqrt{n}})$

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$(15-1.96\cdot\frac{2}{\sqrt{1000}}, 15+1.96\cdot\frac{2}{\sqrt{1000}})\\ \\ (14.88,15.12)$

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A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. The number of subjects needed to estima

Ludmilka [50]

Answer:

$n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65$

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

Step-by-step explanation:

1) Previous concepts

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

ME represent the margin of error

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

2) Solution to the problem

Since the new Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

Assuming that the deviation is known we can express the margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

Replacing into formula (b) we got:

$n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65$

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

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3 years ago