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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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In a triangle the length of two sides are 1.9m and 0.7m. What is the length of the third side, if you know that it should be a w
monitta

Answer:

their sum: 2.8m

their difference: 1.2m

the third side's length should be smaller than their sum, larger than their difference

it could be: 1.3-1.4-1.5-1.6-1.7-1.8-1.9-2-2.1-2.2-2.3-2.4-2.5-2.6-2.7 but since it has to be a whole number, 2m is the only eligible answer.

6 0
2 years ago
Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal pla
svp [43]

Here is  the correct computation of the question given.

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal place. Listed below are the systolic blood pressures (in mm Hg) for a sample of men aged 20-29 and for a sample of men aged 60-69.

Men aged 20-29:      117      122     129      118     131      123

Men aged 60-69:      130     153      141      125    164     139

Group of answer choices

a)

Men aged 20-29: 4.8%

Men aged 60-69: 10.6%

There is substantially more variation in blood pressures of the men aged 60-69.

b)

Men aged 20-29: 4.4%

Men aged 60-69: 8.3%

There is substantially more variation in blood pressures of the men aged 60-69.

c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

d)

Men aged 20-29: 7.6%

Men aged 60-69: 4.7%

There is more variation in blood pressures of the men aged 20-29.

Answer:

(c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

Step-by-step explanation:

From the given question:

The coefficient of variation can be determined by the relation:

coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

We will need to determine the coefficient of variation both men age 20 - 29 and men age 60 -69

To start with;

The coefficient of men age 20 -29

Let's first find the mean and standard deviation before we can do that ;

SO .

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{117+122+129+118+131+123}{6}

Mean = \dfrac{740}{6}

Mean = 123.33

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(117-123.33)^2+(122-123.33)^2+...+(123-123.33)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{161.3334}{5}}

Standard deviation = \sqrt{32.2667}

Standard deviation = 5.68

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{5.68}{123.33}*100

Coefficient of variation = 4.6% for men age 20 -29

For men age 60-69 now;

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{   130 +    153    +  141  +    125 +   164  +   139}{6}

Mean = \dfrac{852}{6}

Mean = 142

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(130-142)^2+(153-142)^2+...+(139-142)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{1048}{5}}

Standard deviation = \sqrt{209.6}

Standard deviation = 14.48

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{14.48}{142}*100

Coefficient of variation = 10.2% for men age 60 - 69

Thus; Option C is correct.

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

4 0
2 years ago
A business school professor uses multiple choice tests. Each question on his exams has five possible answers which are assumed t
avanturin [10]

Answer:

a. The null hypothesis for this test is that the observed distribution is the same as uniform distribution

b. The degrees of freedom do you have for this test is 4

c.  The calculated value of the test statistic is 9.250

Step-by-step explanation:

a. According to the given data we can conclude that the null hypothesis for this test is that the observed distribution is the same as uniform distribution.

b. In order to calculate the degrees of freedom do you have for this test we would have to make the following calculation:

degrees of freedom=k-1

degrees of freedom=5-1

degrees of freedom=4

c. In order to calculate the value of the test statistic first we have to calculate the frecuency expected as follows:

expected frecuency=total observed frecuency/total number of category

expected frecuency=1,000/5

expected frecuency=200

Hence, to calculate the value of the test statistic we have to calculate the following formula:

x∧2=∑(fo-fe)∧2/fe

=(185-200)∧2/200+(230-200)∧2/200+(215-200)∧2/200+(180-200)∧2+(190-200)∧2

=9.250

The calculated value of the test statistic is 9.250

7 0
2 years ago
Using the information given, select the statement that can deduce the line segments to be parallel. If there are none, then sele
Alina [70]

we know that

<u>1) If the line segment AB is parallel to the line segment DC</u>

then

m∠1=m∠D -------> by corresponding angles

<u>2) If the line segment AD is parallel to the line segment BC</u>

then

m∠3=m∠B -------> by corresponding angles

5 0
2 years ago
Read 2 more answers
replace each star with a digit to make the problem true.Is there only one answer to each problem? ****-***=2
Semenov [28]

Answer: We have two solutions:

1000 - 998 = 2

1001 - 999 = 2

Step-by-step explanation:

So we have the problem:

****-*** = 2

where each star is a different digit, so in this case, we have a 4 digit number minus a 3 digit number, and the difference is 2.

we know that if we have a number like 99*, we can add a number between 1 and 9 and we will have a 4-digit as a result:

So we could write this as:

1000 - 998 = 2

now, if we add one to each number, the difference will be the same, and the number of digits in each number will remain equal:

1000 - 998 + 1 - 1 = 2

(1000 + 1) - (998 + 1) = 2

1001 - 999 = 2

now, there is a trivial case where we can find other solutions where the digits can be zero, like:

0004 - 0002 = 2

But this is trivial, so we can ignore this case.

Then we have two different solutions.

7 0
2 years ago
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