Question

According to the article “Modeling and Predicting the Effects of Submerged Arc Weldment Process Parameters on Weldment

Modeling data distributions
ANSWERED
asked 2020-11-24

According to the article “Modeling and Predicting the Effects of Submerged Arc Weldment Process Parameters on Weldment Characteristics and Shape Profiles” (J. of Engr. Manuf., 2012: 1230–1240), the submerged arc welding (SAW) process is commonly used for joining thick plates and pipes. The heat affected zone (HAZ), a band created within the base metal during welding, was of particular interest to the investigators. Here are observations on depth (mm) of the HAZ both when the current setting was high and when it was lower. \(\begin{matrix} Non-high & 1.04 & 1.15 & 1.23 & 1.69 & 1.92 & 1.98 & 2.36 & 2.49 & 2.72 & 1.37 & 1.43 & 1.57 & 1.71 & 1.94 & 2.06 & 2.55 & 2.64 & 2.82 \\ High & 1.55 & 2.02 & 2.02 & 2.05 & 2.35 & 2.57 & 2.93 & 2.94 & 2.97 \\ \end{matrix}\) c. Does it appear that true average HAZ depth is larger for the higher current condition than for the lower condition? Carry out a test of appropriate hypotheses using a significance level of .01.

 

Answers (1)

2020-11-25

Step 1

Given: \(\displaystyle{n}_{{{1}}}={18}\)
\(\displaystyle{n}_{{{2}}}={9}\)
\(\displaystyle\alpha={0.01}\) The mean is the sum of all values divided by the number of values: \(\displaystyle\overline{{{x}}}_{{{1}}}={\frac{{{1.04}+{1.15}+{1.23}+\ldots+{2.55}+{2.64}+{2.82}}}{{{18}}}}\approx{1.9261}\)
\(\displaystyle\overline{{{x}}}_{{{2}}}={\frac{{{1.55}+{2.02}+{2.02}+\ldots+{2.93}+{2.94}+{2.97}}}{{{9}}}}\approx{2.3778}\) The variance is the sum of squared deviations from the mean divided by \(\displaystyle{n}-{1}\). The standard deviation is the square root of the variance: \(\displaystyle{s}_{{{1}}}=\sqrt{{{\frac{{{\left({1.04}-{1.9261}\right)}^{{{2}}}+\ldots.+{\left({2.82}-{1.9261}\right)}^{{{2}}}}}{{{18}-{1}}}}}}\approx{0.5694}\)
\(\displaystyle{s}_{{{2}}}=\sqrt{{{\frac{{{\left({1.55}-{2.3778}\right)}^{{{2}}}+\ldots.+{\left({2.97}-{2.3778}\right)}^{{{2}}}}}{{{9}-{1}}}}}}\approx{0.5072}\)

Given claim: larger the higher current condition. The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain the value mentioned in the claim. \(\displaystyle{H}_{{0}}:\mu_{{{1}}}={u}_{{{2}}}\)
\(\displaystyle{H}_{{\alpha}}:\mu_{{{1}}}{<}\mu_{{{2}}}\)

Step 2

Determine the test statistic: \(\displaystyle{t}={\frac{{\overline{{{x}}}_{{{1}}}-\overline{{{x}}}_{{{2}}}}}{{\sqrt{{{\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{n}_{{{1}}}}}}+{\frac{{{{s}_{{{2}}}^{{{2}}}}}}{{{n}_{{{2}}}}}}}}}}}={\frac{{{1.9261}-{2.3778}}}{{\sqrt{{{\frac{{{0.5694}^{{{2}}}}}{{{18}}}}+{\frac{{{0.5072}^{{{2}}}}}{{{9}}}}}}}}}\approx-{2.093}\)

Determine the degrees of freedom (rounded down to the nearest integer): \(\displaystyle\triangle={\frac{{{\left({\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{n}_{{{1}}}}}}+{\frac{{{{s}_{{{2}}}^{{{2}}}}}}{{{n}_{{{2}}}}}}\right)}}}{{{\frac{{{\left(\frac{{{s}_{{{1}}}^{{{2}}}}}{{n}_{{{1}}}}\right)}^{{{2}}}}}{{{n}_{{{1}}}-{1}}}}+{\frac{{{\left(\frac{{{s}_{{{2}}}^{{{2}}}}}{{n}_{{{2}}}}\right)}^{{{2}}}}}{{{n}_{{{2}}}-{1}}}}}}}={\frac{{{\left({\frac{{{0.5694}^{{{2}}}}}{{{18}}}}+{\frac{{{0.5072}^{{{2}}}}}{{{9}}}}\right)}^{{{2}}}}}{{{\frac{{{\left(\frac{{0.5694}^{{{2}}}}{{18}}\right)}^{{{2}}}}}{{{18}-{1}}}}+{\frac{{{\left(\frac{{0.5072}^{{{2}}}}{{9}}\right)}^{{{2}}}}}{{{9}-{1}}}}}}}\approx{17}\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Student's T distribution in the appendix containing the t-value in the \(\displaystyle{d}{f}={17}\): \(\displaystyle{0.025}{<}{P}{<}{0.05}\)

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected: \(\displaystyle{P}{>}{0.01}\Rightarrow\) Fail to reject \(\displaystyle{H}_{{{0}}}\) There is not sufficient evidence to support the claim that the true average HAZ depth is larger for the higher current condition than for the lower condition.

0
 
Best answer

expert advice

Need a better answer?

Relevant Questions

asked 2021-01-23

The article “Stochastic Modeling for Pavement Warranty Cost Estimation” (J. of Constr. Engr. and Mgmnt., 2009: 352–359) proposes the following model for the distribution of Y = time to pavement failure. Let \(\displaystyle{X}_{{{1}}}\) be the time to failure due to rutting, and \(\displaystyle{X}_{{{2}}}\) be the time to failure due to transverse cracking, these two rvs are assumed independent. Then \(\displaystyle{Y}=\min{\left({X}_{{{1}}},{X}_{{{2}}}\right)}\). The probability of failure due to either one of these distress modes is assumed to be an increasing function of time t. After making certain distributional assumptions, the following form of the cdf for each mode is obtained: \(\displaystyle\Phi{\left[\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right]}\) where \(\Uparrow \Phi\) is the standard normal cdf. Values of the five parameters a, b, c, d, and e are -25.49, 1.15, 4.45, -1.78, and .171 for cracking and -21.27, .0325, .972, -.00028, and .00022 for rutting. Determine the probability of pavement failure within \(\displaystyle{t}={5}\) years and also \(\displaystyle{t}={10}\) years.

asked 2021-03-07

In an experiment designed to study the effects of illumination level on task performance (“Performance of Complex Tasks Under Different Levels of Illumination,” J. Illuminating Eng., 1976: 235–242), subjects were required to insert a fine-tipped probe into the eyeholes of ten needles in rapid succession both for a low light level with a black background and a higher level with a white background. Each data value is the time (sec) required to complete the task.
\(\begin{array}{|c|c|} \hline Subject & (1) & (2) & (3) & (4) & (5) &(6) & (7) & (8) & (9) \\ \hline Black & 25.85 & 28.84 & 32.05 & 25.74 & 20.89 & 41.05 & 25.01 & 24.96 & 27.47 \\ \hline White & 18.28 & 20.84 & 22.96 & 19.68 & 19.509 & 24.98 & 16.61 & 16.07 & 24.59 \\ \hline \end{array}\)
Does the data indicate that the higher level of illumination yields a decrease of more than 5 sec in true average task completion time? Test the appropriate hypotheses using the P-value approach.

asked 2020-11-26

An analysis of laboratory data collected with the goal of modeling the weight (in grams) of a bacterial culture after several hours of growth produced the least squares regression line \(\log(weight) = 0.25 + 0.61\)hours. Estimate the weight of the culture after 3 hours.

A) 0.32 g

B) 2.08 g

C) 8.0 g

D) 67.9 g

E) 120.2 g

asked 2020-12-07

Would you rather spend more federal taxes on art? Of a random sample of \(n_{1} = 86\) politically conservative voters, \(r_{1} = 18\) responded yes. Another random sample of \(n_{2} = 85\) politically moderate voters showed that \(r_{2} = 21\) responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use \(\alpha = 0.05.\)

a) State the null and alternate hypotheses.

 

\(H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2\)

\(H_0:p_{1} = p_{2}, H_{1}:p_{1} < p_2\)

\(H_0:p_{1} = p_{2}, H_{1}:p_{1} \neq p_2\)

\(H_{0}:p_{1} < p_{2}, H_{1}:p_{1} = p_{2}\)

b) What sampling distribution will you use? What assumptions are you making? The Student's t. The number of trials is sufficiently large. The standard normal. The number of trials is sufficiently large.The standard normal. We assume the population distributions are approximately normal. The Student's t. We assume the population distributions are approximately normal.

c)What is the value of the sample test statistic? (Test the difference \(p_{1} - p_{2}\). Do not use rounded values. Round your final answer to two decimal places.)

d) Find (or estimate) the P-value. (Round your answer to four decimal places.)

e) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level alpha? At the \(\alpha = 0.05\) level, we reject the null hypothesis and conclude the data are statistically significant. At the \(\alpha = 0.05\) level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the \(\alpha = 0.05\) level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the \(\alpha = 0.05\) level, we reject the null hypothesis and conclude the data are not statistically significant.

f) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters.

asked 2021-03-02

An experiment designed to study the relationship between hypertension and cigarette smoking yielded the following data.
\(\begin{array}{|c|c|} \hline Tension\ level & Non-smoker & Moderate\ smoker & Heavy\ smoker \\ \hline Hypertension & 20 & 38 & 28 \\ \hline No\ hypertension & 50 & 27 & 18 \\ \hline \end{array}\)
Test the hypothesis that whether or not an individual has hypertension is independent of how much that person smokes.

asked 2021-03-02

The following observations are lifetimes (days) subsequent to diagnosis for individuals suffering from blood cancer ("A Goodness of Fit Approach to the Class of Life Distributions with Unknown Age," Quality and Reliability Engr. Intl., \(2012: 761-766): 115, 181, 255, 418, 441, 461, 516, 739, 743, 789, 807, 865, 924, 983, 1025, 1062, 1063, 1165, 1191, 1222, 1222, 1251, 1277, 1290, 1357, 1369, 1408, 1455, 1278, 1519, 1578, 1578, 1599, 1603, 1605, 1696, 1735, 1799, 1815, 1852, 1899, 1925, 1965.\)
a) can a confidence interval for true average lifetime be calculated without assuming anything about the nature of the lifetime distribution? Explain your reasoning. [Note: A normal probability plot of data exhibits a reasonably linear pattern.]
b) Calculate and interpret a confidence interval with a 99% confidence level for true average lifetime. [Hint: mean \(= 1191.6, s = 506.6\).]

asked 2020-11-07

1)A rewiew of voted registration record in a small town yielded the dollowing data of the number of males and females registered as Democrat, Republican, or some other affilation:

\(\begin{array}{c} Gender \\ \hline Affilation & Male & Female \\ \hline Democrat & 300 & 600 \\ Republican & 500 & 300 \\ Other & 200 & 100 \\ \hline \end{array}\)

What proportion of all voters is male and registered as a Democrat? 2)A survey was conducted invocted involving 303 subject concerning their preferences with respect to the size of car thay would consider purchasing. The following table shows the count of the responses by gender of the respondents:

\(\begin{array}{c} Size\ of\ Car \\ \hline Gender & Small & Medium & lange & Total \\ \hline Female & 58 & 63 & 17 & 138 \\ Male & 79 & 61 & 25 & 165 \\ Total & 137 & 124 & 42 & 303 \\ \hline \end{array}\)

the data are to be summarized by constructing marginal distributions. In the marginal distributio for car size, the entry for mediums car is ?

asked 2020-12-28

The table shows the population of various cities, in thousands, and the average walking speed, in feet per second, of a person living in the city. \(\begin{array}{|c|c|} \hline Population\ (thousands) & Walking Speed\ (feet\ per\ second) \\ \hline 5.5 & 0.6 \\ \hline 14 & 1.0\\ \hline 71 & 1.6\\ \hline 138 & 1.9\\ \hline 342 & 2.2\\ \hline \end{array}\)

asked 2020-12-30
The tables show the battery lives (in hours) of two brands of laptops. a) Make a double box-and-whisker plot that represent's the data. b) Identifity the shape of each distribution. c) Which brand's battery lives are more spread out? Explain. d) Compare the distributions using their shapes and appropriate measures of center and variation.
asked 2021-02-02

Determine which of the following functions \(\displaystyle{f{{\left({x}\right)}}}={c}{x},\ {g{{\left({x}\right)}}}={c}{x}^{{{2}}},\ {h}{\left({x}\right)}={c}\sqrt{{{\left|{x}\right|}}},\ \text{and}\ {r}{\left({x}\right)}=\ {\frac{{{c}}}{{{x}}}}\) can be used to model the data and determine the value of the constant c that will make the function fit the data in the table. \(\begin{array}{|c|c|} \hline x & -4 & -1 & 0 & 1 & 4 \\ \hline y & -32 & -2 & 0 & -2 & -32 \\ \hline \end {array}\)

You might be interested in

...