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igomit [66]
2 years ago
10

A herd of 1,500 steer was fed a special high‐protein grain for a month. A random sample of 29 were weighed and had gained an ave

rage of 6.7 pounds. If the standard deviation of weight gain for the entire herd is 7.1, test the hypothesis that the average weight gain per steer for the month was more than 5 pounds.

Mathematics
1 answer:
Marianna [84]2 years ago
4 0

By formula we know that:

z (x) = (x - m) / [sd / sqrt (n)]

where x is the value we want to know (6.7), m is the mean (5), sd is the standard deviation (7.1) and n is the sample size (29).

Replacing we have:

z (6.7) = (6.7 - 5) / [7.1 / sqrt (29)]

z = 1.289

If we look in the normal distribution table (attached), we have that the probability is 0.8997, therefore:

1 - 0.8897 = 0.1003

So the conditional probability of a herd sample earning at least 6.7 pounds per steer is 10.03%.

Now the hypothesis tells me:

m> 5

The probability is somewhat low, therefore, the most correct thing is to reject the hypothesis even though it is a fact that can occur.

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An engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation is 1.6
nexus9112 [7]

Answer:

n=1705

Step-by-step explanation:

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".  

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma=1.6)  

And the distribution for \bar X is:

\bar X \sim N(\mu, \frac{1.6}{\sqrt{n}})  

We know that the margin of error for a confidence interval is given by:  

Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)  

The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.99=0.01 and \alpha/2=0.005  

Using the normal standard table, excel or a calculator we see that:  

z_{\alpha/2}=\pm 2.58  

If we solve for n from formula (1) we got:  

\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}  

n=(\frac{z_{\alpha/2} \sigma}{Me})^2  

And we have everything to replace into the formula:  

n=(\frac{2.58(1.6)}{0.1})^2 =1704.03  

And if we round up the answer we see that the value of n to ensure the margin of error required \pm=0.1 mm is n=1705.  

6 0
2 years ago
Can u help me <br><br>thanks u
vovikov84 [41]
1 1/2=
1.5

1.5 * 3 =

4.5
7 0
2 years ago
I need help with #39-42, please explain.
shusha [124]

For question 39 & 40, we need to use the below equation to complete the sentence

l=m\sqrt{n}

Question 39:

When ' n ' increase, the \sqrt{n} will also increase and that multiplied with constant ' m ', the l will also increase.

Solution for question 39:

As n increases and m stays constant , l <u>increases</u>

-------

Question 40:

Solving the equation for m, we get

l = m\sqrt{n} \\ \\ m=\frac{l}{\sqrt{n}}

When ' l ' increases, the numerator increase, the denominator stays constant because 'n' stays constant, for this condition, the fraction increases.

Solution for question 40:

As l increase and n stays constant, m <em><u>increases</u></em>

------

For question 41 & 42, we need to use the below equation to complete the sentence

r=s^2/t^2

Question 41:

When s is triped, the equation will be...

r=(3s)^2/t^2=\frac{3^{2}s^{2}}{t^2}   =9s^2/t^2

Solution for question 41:

If s is tripled and t stays constant, r is multiplied by <em><u>9</u></em>

--------

Question 42:

When t is doubled, the equation will be...

r=s^2/(2t)^2=\frac{s^2}{2^2 \cdot t^2}=\frac{s^2}{4t^2}\\   \\ r=0.25s^2/t^2 \; \; (or) \; \; \frac{1}{4} \cdot \frac{s^2}{t^2}

Solution for 42:

If t doubled and s stays constant, r is multiplied by <em><u>1/4 or 0.25</u></em>


3 0
3 years ago
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