 Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type. \begin{array}{|c|c|}\hline 11.8 & 7.7 & 6.5 & 6 .8& 9. jernplate8 2021-05-14 Answered
Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type.
$$\begin{array}{|c|c|}\hline 11.8 & 7.7 & 6.5 & 6 .8& 9.7 & 6.8 & 7.3 \\ \hline 7.9 & 9.7 & 8.7 & 8.1 & 8.5 & 6.3 & 7.0 \\ \hline 7.3 & 7.4 & 5.3 & 9.0 & 8.1 & 11.3 & 6.3 \\ \hline 7.2 & 7.7 & 7.8 & 11.6 & 10.7 & 7.0 \\ \hline \end{array}$$
a) Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion. $$[Hint.\ ?x_{j}=219.5.]$$ (Round your answer to three decimal places.)
MPa
State which estimator you used.
$$x$$
$$p?$$
$$\frac{s}{x}$$
$$s$$
$$\tilde{\chi}$$
b) Calculate a point estimate of the strength value that separates the weakest $$50\%$$ of all such beams from the strongest $$50\%$$.
MPa
State which estimator you used.
$$s$$
$$x$$
$$p?$$
$$\tilde{\chi}$$
$$\frac{s}{x}$$
c) Calculate a point estimate of the population standard deviation ?. $$[Hint:\ ?x_{i}2 = 1859.53.]$$ (Round your answer to three decimal places.)
MPa
Interpret this point estimate.
This estimate describes the linearity of the data.
This estimate describes the bias of the data.
This estimate describes the spread of the data.
This estimate describes the center of the data.
Which estimator did you use?
$$\tilde{\chi}$$
$$x$$
$$s$$
$$\frac{s}{x}$$
$$p?$$
d) Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa. [Hint: Think of an observation as a "success" if it exceeds 10.] (Round your answer to three decimal places.)
e) Calculate a point estimate of the population coefficient of variation $$\frac{?}{?}$$. (Round your answer to four decimal places.)
State which estimator you used.
$$p?$$
$$\tilde{\chi}$$
$$s$$
$$\frac{s}{x}$$
$$x$$

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Step 1
The point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion is, $$\mu$$
$$\hat{\mu}=\bar{x}$$
$$=\frac{\sum x_{i}}{n}$$
$$=\frac{219.5}{27}$$
$$=8.12963$$
The estimator is $$\bar{X}$$
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 $$\bar{X}$$
Step 2
First arrange the data into ascending order.
$$\begin{array}{|c|c|}\hline A & B & C \\ \hline 5.3 & & 8.1 \\ \hline 6.3 & & 8.1 \\ \hline 6.3 & & 8.1 \\ \hline 6.5 & & 8.5 \\ \hline 6.8 & & 8.7 \\ \hline 6.8 & & 9 \\ \hline 7 & & 9.7 \\ \hline 7 & & 9.7 \\ \hline 7.2 & & 10.7 \\ \hline 7.3 & & 11.3 \\ \hline 7.3 & & 11.6 \\ \hline 7.4 & & 11.8 \\ \hline 7.7 & & \\ \hline 7.7 & & \\ \hline 7.8 & & \\ \hline \end{array}$$
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 $$x$$
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 $$x$$
Step 3
The point estimate of the population standard deviation is,
$$\hat{\sigma}=s=\sqrt{\frac{\sum x_{i}^{2}-(\sum x_{i})^{2}/n}{n-1}}$$
$$=\sqrt{\frac{1859.53-(219.5)^{2}/27}{27-1}}$$
$$=\sqrt{\frac{1859.53-1784.454}{26}}$$
$$s=1.699$$
The estimator is $$s$$
Part c
The point estimate of the population standard deviation is 1.699. The estimator is $$s$$
Step 4
Let $$x$$ 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,
$$\hat{p}=\frac{x}{n}$$
$$=\frac{4}{27}$$
$$=0.148$$
Part d
The point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa is 0.148.
Step 5
The point estimate of the proportion coefficient of variation is,
$$\frac{\sigma}{\mu}=\frac{S}{\bar{x}}\times100$$
$$=\frac{1.699}{8.12963}\times100$$
$$=20.9$$
The estimator is,
$$\frac{\sigma}{\mu}=\frac{S}{\bar{x}}$$
Part e
The point estimate of the proportion coefficient of variation is 20.9. The estimator is $$\frac{s}{\bar{x}}$$
Coefficient variation is the ratio of the standard deviation and mean and multiply with 100.
Not exactly what you’re looking for? content_user

a.The point estimate of the mean value is the sample mean. You calculate it using the following formula:

$$X[bar]= \sum \frac{X}{n}= \frac{219.40}{27}= 8.1259 \cong 8.13$$

b.

The point estimate that divides the sample in 50%-50% is the measure of central tendency called Median(Me).

To reach the median you have to calculate it's position (PosMe)

For uneven samples PosMe= $$\frac{(n+1)}{2}= \frac{(27+1)}{2}= 14 \Rightarrow$$ This means that the Median is the 14th observation of the data set.

So next is to order the data from lower to heighest and identify the value in the 14th position

c.

The point estimate of the population standard deviation is the sample standard deviation. The standard deviation is the square root of the variance, so the first step is to calculate the sample variance:

$$S^2=\frac{1}{n-1} \cdot [sum X^2-\frac{(sum X)^2}{n}]$$

$$\sum X^2=1858.92$$

$$S^2=\frac{1}{26} \cdot [1858.92-\frac{219.40^2}{27}]=2.926$$

$$S=\sqrt{2.926}=1.71$$

The standard deviation is a measurement of variability, it shows how to disperse the data is concerning the sample mean.

d. This estimate describes the spread of the data.