Consider the accompanying data on flexural strength (MPa) forconcrete beams of a certain type.

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a) Calculate a point estimate of the mean value of strength forthe conceptual population of all beams manufactured in thisfashion. <Hint: ?xi = 219.5.> (Roundyour answer to three decimal places.) MPa

State which estimator you used.

x

p?


s / x

s

(b) Calculate a point estimate of the strength value that separatesthe weakest 50% of all such beams from the strongest 50%. MPa

State which estimator you used.


s

x

p?

s / x

(c) Calculate a point estimate of the population standard deviation?. <Hint: ?xi2 =1859.53.> (Round your answer to three decimal places.) MPa

Interpret this point estimate.

This estimate describes the linearity of the data. This estimatedescribes the bias of the data. Thisestimate describes the spread of the data. This estimate describesthe center of the data.

Which estimator did you use?

x

s

s / x

p?

(d) Calculate a point estimate of the proportion of all such beamswhose flexural strength exceeds 10 MPa. <Hint: Think of anobservation as a "success" if it exceeds 10.> (Round your answer tothree decimal places.)

(e) Calculate a point estimate of the population coefficient ofvariation ?/?. (Round your answer to four decimalplaces.)

State which estimator you used.

p?

s

s / x

x

Answer


General guidance


Concepts and reason
Mean is ratio of the sum of the observations and the number of observations.

Median is the middle value in the list of numbers. To find the median, first list the numbers from smallest to largest. Identify the given numbers; if the given numbers are odd then the median is the middle term. If the given numbers are even the median is the average of the median of the two terms.

Standard deviation can be calculated by the obtaining the square root of the variance and its value cannot be negative.


Coefficient variation is the ratio of the standard deviation and mean. It is used to identify which data set is more consistent among the two data sets.

Sample mean is an unbiased estimate of the population mean.


Fundamentals

The formula for mean is,

*

The formula for median is,

*

The formula for standard deviation is,

*

The formula for coefficient of variation is,

The estimator and estimate are shown in the below table:


Step-by-step


Step 1 of 5


(a)

The point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion is,


The point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion is

*
.

*

The estimator is


Part a

The point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion is 8.12963. The estimator is


The mean of the each sample can be obtained by ration of the sum of the observations and number of observations.


Use middle value to find the point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50% MPa.


Step 2 of 5


(b)

First arrange the data into ascending order.

*

Here, the number of observations is odd. So, the median is the middle value of the of the given data.

Hence, the middle value is 7.7.

The estimator is

*


Part b

The point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50% MPa is 7.7. The estimator is


The median is separates the weakest 50% of all such beams from the strongest 50% MPa.


Use

*
to find the point estimate of the population standard deviation.


Step 3 of 5


(c)

The point estimate of the population standard deviation is,

*

The estimator is


Part c

The point estimate of the population standard deviation is 1.699. The estimator is


The point estimate of the population standard deviation describes the spread of the data.


Use

*
to find the point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPA.


Step 4 of 5


(d)

Let denote the number all such beams whose flexural strength exceeds 10 MPA.

From the given information number of beams whose flexural strength exceeds 10 MPA is 4.

The point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa is,


Part d

The point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa is 0.148.


The proportion is the ratio of the favorable number of cases divided by total number of cases.


Use

*
to find the point estimate of the proportion coefficient of variation.


Step 5 of 5


(e)

The point estimate of the proportion coefficient of variation is,

*

The estimator is,

*


Part e

The point estimate of the proportion coefficient of variation is 20.9. The estimator is


Coefficient variation is the ratio of the standard deviation and mean and multiply with 100.


Answer


Part a

The point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion is 8.12963. The estimator is


Part b

The point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50% MPa is 7.7. The estimator is


Part c

The point estimate of the population standard deviation is 1.699. The estimator is


Part d

The point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa is 0.148.


Part e

The point estimate of the proportion coefficient of variation is 20.9. The estimator is


Answer only


Part a

The point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion is 8.12963. The estimator is


Part b

The point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50% MPa is 7.7. The estimator is


Part c

The point estimate of the population standard deviation is 1.699. The estimator is


Part d

The point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa is 0.148.

See more: Penn &Amp; Teller: Fool Us Season 5 Episode 1, University Of Pennsylvania


Part e

The point estimate of the proportion coefficient of variation is 20.9. The estimator is


x =
observation (If number of observation is odd) Median=1 observation (If number of observation is odd)
5 = Σx - (Σκ) η η-1
001 = 40
Estimator Estimate Σ * * 6 Ex? - (2x;)in n-1 615
û= 219.5 27 = 8.12963
А в с 5.3 7.9| 6.3 8.1 6.3 8.1| 6.5 8.5| | 6. 8 8 .7 6.8 79.7 7 9.7 7.2 10.7 7.3 11.3 | 11.6 7.4 11.8 7.7 7.8
5 = Σx - (Σκ) η η-1
= 0.148
си
= -x100 1.699 8.12963 = 20.9
| 6 | 3
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