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.
\(p?\)
\(\tilde{\chi}\)
\(s\)
\(\frac{s}{x}\)
\(x\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

liingliing8
Answered 2021-05-15 Author has 21292 answers
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?
Ask My Question
23
 
content_user
Answered 2021-10-12 Author has 11052 answers

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.

16

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-05-21
The pmf of the amount of memory X (GB) in a purchased flash drive is given as the following.
\(\begin{array}{|c|c|}\hline x & 1 & 2 & 4 & 8 & 16 \\ \hline p(x) & 0.05 & 0.10 & 0.30 & 0.45 & 0.10 \\ \hline \end{array} \)
a) Compute E(X). (Enter your answer to two decimal places.) GB
b) Compute V(X) directly from the definition. (Enter your answer to four decimal places.) \(GB^{2}\)
c) Compute the standard deviation of X. (Round your answer to three decimal places.) GB
d) Compute V(X) using the shortcut formula. (Enter your answer to four decimal places.) \(GB^{2}\)
asked 2021-09-18

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. \(\displaystyle{\left[{H}\int.\ ?{x}_{{{j}}}={219.5}.\right]}\) (Round your answer to three decimal places.)
MPa
State which estimator you used.
\(\displaystyle{x}\)
\(\displaystyle{p}?\)
\(\displaystyle{\frac{{{s}}}{{{x}}}}\)
\(\displaystyle{s}\)
\(\displaystyle\tilde{{\chi}}\)
b) Calculate a point estimate of the strength value that separates the weakest \(\displaystyle{50}\%\) of all such beams from the strongest \(\displaystyle{50}\%\).
MPa
State which estimator you used.
\(\displaystyle{s}\)
\(\displaystyle{x}\)
\(\displaystyle{p}?\)
\(\displaystyle\tilde{{\chi}}\)
\(\displaystyle{\frac{{{s}}}{{{x}}}}\)
c) Calculate a point estimate of the population standard deviation ?. \(\displaystyle{\left[{H}\int:\ ?{x}_{{{i}}}{2}={1859.53}.\right]}\) (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?
\(\displaystyle\tilde{{\chi}}\)
\(\displaystyle{x}\)
\(\displaystyle{s}\)
\(\displaystyle{\frac{{{s}}}{{{x}}}}\)
\(\displaystyle{p}?\)
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 \(\displaystyle{\frac{{?}}{{?}}}\). (Round your answer to four decimal places.)
State which estimator you used.
\(\displaystyle{p}?\)
\(\displaystyle\tilde{{\chi}}\)
\(\displaystyle{s}\)
\(\displaystyle{\frac{{{s}}}{{{x}}}}\)
\(\displaystyle{x}\)

asked 2021-06-13
1. Who seems to have more variability in their shoe sizes, men or women?
a) Men
b) Women
c) Neither group show variability
d) Flag this Question
2. In general, why use the estimate of \(n-1\) rather than n in the computation of the standard deviation and variance?
a) The estimate n-1 is better because it is used for calculating the population variance and standard deviation
b) The estimate n-1 is never used to calculate the sample variance and standard deviation
c) \(n-1\) provides an unbiased estimate of the population and allows more variability when using a sample and gives a better mathematical estimate of the population
d) The estimate n-1 is better because it is use for calculation of both the population and sample variance as well as standard deviation.
\(\begin{array}{|c|c|}\hline \text{Shoe Size (in cm)} & \text{Gender (M of F)} \\ \hline 25.7 & M \\ \hline 25.4 & F \\ \hline 23.8 & F \\ \hline 25.4 & F \\ \hline 26.7 & M \\ \hline 23.8 & F \\ \hline 25.4 & F \\ \hline 25.4 & F \\ \hline 25.7 & M \\ \hline 25.7 & F \\ \hline 23.5 & F \\ \hline 23.1 & F \\ \hline 26 & M \\ \hline 23.5 & F \\ \hline 26.7 & F \\ \hline 26 & M \\ \hline 23.1 & F \\ \hline 25.1 & F \\ \hline 27 & M \\ \hline 25.4 & F \\ \hline 23.5 & F \\ \hline 23.8 & F \\ \hline 27 & M \\ \hline 25.7 & F \\ \hline \end{array}\)
\(\begin{array}{|c|c|}\hline \text{Shoe Size (in cm)} & \text{Gender (M of F)} \\ \hline 27.6 & M \\ \hline 26.9 & F \\ \hline 26 & F \\ \hline 28.4 & M \\ \hline 23.5 & F \\ \hline 27 & F \\ \hline 25.1 & F \\ \hline 28.4 & M \\ \hline 23.1 & F \\ \hline 23.8 & F \\ \hline 26 & F \\ \hline 25.4 & M \\ \hline 23.8 & F \\ \hline 24.8 & M \\ \hline 25.1 & F \\ \hline 24.8 & F \\ \hline 26 & M \\ \hline 25.4 & F \\ \hline 26 & M \\ \hline 27 & M \\ \hline 25.7 & F \\ \hline 27 & M \\ \hline 23.5 & F \\ \hline 29 & F \\ \hline \end{array}\)
asked 2021-05-14
When σ is unknown and the sample size is \(\displaystyle{n}\geq{30}\), there are tow methods for computing confidence intervals for μμ. Method 1: Use the Student's t distribution with d.f. = n - 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When \(\displaystyle{n}\geq{30}\), use the sample standard deviation s as an estimate for σσ, and then use the standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for σσ. Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution. Consider a random sample of size n = 31, with sample mean x¯=45.2 and sample standard deviation s = 5.3. (c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?
asked 2021-06-05
Use the following Normal Distribution table to calculate the area under the Normal Curve (Shaded area in the Figure) when \(Z=1.3\) and \(H=0.05\);
Assume that you do not have vales of the area beyond \(z=1.2\) in the table; i.e. you may need to use the extrapolation.
Check your calculated value and compare with the values in the table \([for\ z=1.3\ and\ H=0.05]\).
Calculate your percentage of error in the estimation.
How do I solve this problem using extrapolation?
\(\begin{array}{|c|c|}\hline Z+H & Prob. & Extrapolation \\ \hline 1.20000 & 0.38490 & Differences \\ \hline 1.21000 & 0.38690 & 0.00200 \\ \hline 1.22000 & 0.38880 & 0.00190 \\ \hline 1.23000 & 0.39070 & 0.00190 \\ \hline 1.24000 & 0.39250 & 0.00180 \\ \hline 1.25000 & 0.39440 & 0.00190 \\ \hline 1.26000 & 0.39620 & 0.00180 \\ \hline 1.27000 & 0.39800 & 0.00180 \\ \hline 1.28000 & 0.39970 & 0.00170 \\ \hline 1.29000 & 0.40150 & 0.00180 \\ \hline 1.30000 & 0.40320 & 0.00170 \\ \hline 1.31000 & 0.40490 & 0.00170 \\ \hline 1.32000 & 0.40660 & 0.00170 \\ \hline 1.33000 & 0.40830 & 0.00170 \\ \hline 1.34000 & 0.41010 & 0.00180 \\ \hline 1.35000 & 0.41190 & 0.00180 \\ \hline \end{array}\)
asked 2021-08-01

For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.
\(\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline f(x) & 20 & 21.6 & 29.2 & 36.4 & 46.6 & 55.7 & 72.6 & 87.1 & 107.2 & 138.1 \\ \hline \end{array}\)

asked 2021-06-09
Use the table of values of \(f(x, y)\) to estimate the values of \(fx(3, 2)\), \(fx(3, 2.2)\), and \(fxy(3, 2)\).
\(\begin{array}{|c|c|}\hline y & 1.8 & 2.0 & 2.2 \\ \hline x & & & \\ \hline 2.5 & 12.5 & 10.2 & 9.3 \\ \hline 3.0 & 18.1 & 17.5 & 15.9 \\ \hline 3.5 & 20.0 & 22.4 & 26.1 \\ \hline \end{array}\)
...