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Aleonysh [2.5K]
2 years ago
7

The manufacturer of hardness testing equipment uses steel-ball indenters to penetrate metal that is being tested, however, the m

anufacturer thinks it would be better to use a diamond indenter so that all ypes of metal can be tested. Because of differences between the two types of indenters, it is suspected that the two methods will produce different hardness readings. The metal speciments to be tested are large enough so that two indentions can be made. Therefore, the manufacturer uses both indenters on each specimen and compares the hardness readings. Construct a 95% confidence interval to judge whether the two indenters resul in different measurements. Note: A normal probability plot and boxplot indicate that the differences are approximately normally distributed with no outliers.Specimen Steel ball Diamond1 51 532 57 553 61 634 70 745 68 696 54 567 65 688 51 519 53 56Construct a 95% confidence interval to judge whether the two indenters result in different measurements, where the differences are computed as "diamond minus steel ball"The 95% confidence interval to judge whether the two indeners result in different measurements is?
Mathematics
1 answer:
masha68 [24]2 years ago
4 0

Answer:

1.667-2.31\frac{1.803}{\sqrt{9}}=0.2789  

1.667+2.31\frac{1.803}{\sqrt{9}}=3.055  

So on this case the 95% confidence interval would be given by (0.2789;3.055)  

The 95% confidence interval to judge whether the two indeners result in different measurements is?

Yes the confidence interval not contains the value 0 so we can conclude that the values for Diamond are significantly higher than the values for Steel Ball at 5% of significance.

Step-by-step explanation:

We have the following dataset:

specimen    1     2    3     4      5     6     7    8     9

Steel Ball   51   57   61   70   68   54   65  51   53

Diamond   53   55  63   74   69   56   68  51   56

If we calculate the differences diamond-steel ball we have this datase:

d: 2, -2, 2, 4, 1, 2, 3, 0, 3

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}=1.667

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1.803

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

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

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n-1=9-1=8  

Since the confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that t_{\alpha/2}=2.31.

Now we have everything in order to replace into formula (1):  

1.667-2.31\frac{1.803}{\sqrt{9}}=0.2789  

1.667+2.31\frac{1.803}{\sqrt{9}}=3.055  

So on this case the 95% confidence interval would be given by (0.2789;3.055)  

The 95% confidence interval to judge whether the two indeners result in different measurements is?

Yes the confidence interval not contains the value 0 so we can conclude that the values for Diamond are significantly higher than the values for Steel Ball at 5% of significance.

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Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 t
musickatia [10]

Answer:

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

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

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Step-by-step explanation:

Assuming this complete question: Which statement gives a valid interpretation of the interval?

The store manager is 90% confident that the average amount spent by the 36 sampled customers is between S38.01 and $44.29.

There is a 90% chance that the mean amount spent by all customers is between S38.01 and $44.29.

There is a 90% chance that a randomly selected customer will spend between S38.01 and $44.29.

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

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".

Solution to the problem

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

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

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

8 0
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Initially 5 grams of salt are dissolved into 10 liters of water. Brine with concentration of salt 5 grams per liter is added at
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Salt flows in at a rate of (5 g/L)*(3 L/min) = 15 g/min.

Salt flows out at a rate of (x/10 g/L)*(3 L/min) = 3x/10 g/min.

So the net flow rate of salt, given by x(t) in grams, is governed by the differential equation,

x'(t)=15-\dfrac{3x(t)}{10}

which is linear. Move the x term to the right side, then multiply both sides by e^{3t/10}:

e^{3t/10}x'+\dfrac{3e^{t/10}}{10}x=15e^{3t/10}

\implies\left(e^{3t/10}x\right)'=15e^{3t/10}

Integrate both sides, then solve for x:

e^{3t/10}x=50e^{3t/10}+C

\implies x(t)=50+Ce^{-3t/10}

Since the tank starts with 5 g of salt at time t=0, we have

5=50+C\implies C=-45

\implies\boxed{x(t)=50-45e^{-3t/10}}

The time it takes for the tank to hold 20 g of salt is t such that

20=50-45e^{-3t/10}\implies t=\dfrac{20}3\ln\dfrac32\approx2.7031\,\mathrm{min}

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the richmond park gamekeeper wanted to know how many deer were in the park . he caught 90 of them and put a small tag round a ho
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Answer:

a. Since 10 out of 70 had a tag we can write the ratio 10:70. We need to solve for x in 90:x. x = 630 so the answer is 630.

b. An assumption he could have made is that the number of deer that had a tag and the total amount of deer were directly proportional.

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The graph shows the number of each kind of book in Hero's personal library. A book is chosen randomly. What is the probability t
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14 science fiction, 2 romance, 4 biography, 10 adventure...total of 30 books

P(biography) = biographies / total books = 4/30 which reduces to 2/15 <=
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Answer:

answer is C

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