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

Question
Modeling data distributions
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. PSK\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}ZSK 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. \(\displaystyle{H}_{{0}}:\mu_{{{1}}}={u}_{{{2}}}\)
\(\displaystyle{H}_{{\alpha}}:\mu_{{{1}}}{<}\mu_{{{2}}}\)</span> 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 \(\displaystyle{d}{f}={17}\): \(\displaystyle{0.025}{<}{P}{<}{0.05}\)</span> 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.
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State the null and alternate hypotheses.
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\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}\)
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The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
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What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimal places.)
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\(\displaystyle{0},{005}<{P}-\text{value}<{0},{025}\)
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A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of \(25^{\circ}F\). However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to \(25^{\circ}F\). One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a \(5\%\) level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
\(H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}\)
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
\(df_{N} = ?\)
\(df_{D} = ?\)
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
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