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Select a property of the F distribution


A) It is symmetric.
B) The exact shape of the F distribution depends on two different degrees of freedom.
C) Values of the F distribution can be negative.
D) The total area under the curve is equal to 2.

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Given below are the analysis of variance results from a Minitab display. to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. Identify the value of the test statistic.  Source DFSSMSFp Factor 313.5004.5005.170.011 Error 1613.9250.870 Total 1927.425\begin{array} { l | r | c | c | c | c } \text { Source } & \boldsymbol { D F } & \boldsymbol { S S } & \boldsymbol { M S } & \boldsymbol { F } & \boldsymbol { p } \\\hline \text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\\hline \text { Error } & 16 & 13.925 & 0.870 & & \\\hline \text { Total } & 19 & 27.425 & & &\end{array}


A) 5.17
B) 4.500
C) 13.500
D) 0.011

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The following data contains task completion times, in minutes, categorized according to the gender of the machine operator and the machine used.  Male  Female  Machine 1 15,1716,17 Machine 2 14,1315,13 Machine 3 16,1817,19\begin{array} { l c c } & \text { Male } & \text { Female } \\\text { Machine 1 } & 15,17 & 16,17 \\\text { Machine 2 } & 14,13 & 15,13 \\\text { Machine 3 } & 16,18 & 17,19\end{array} The ANOVA results lead us to conclude that the completion times are not affected by an interaction between machine and gender, and the times are not affected by gender, but they are affected by the machine. Change the table entries so that there is no effect from the interaction between machine and gender, but there is an effect from the gender of the operator.

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The following table is one example of en...

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Which of the following is NOT a requirement for one-way ANOVA?


A) The populations have distributions that are approximately normal.
B) The samples are simple random samples.
C) The samples are independent of each other.
D) The data is categorical data.

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Explain the procedure for two-way analysis of variance, and how it varies depending on whether there is an interaction between the two factors or not.

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First we test whether there is an effect...

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The following data shows the yield, in bushels per acre, categorized according to three varieties of corn and three different soil conditions. Assume that yields are not affected by an interaction between variety and soil conditions, and test the null hypothesis that variety has no effect on yield. Use a 0.05 significance level.  Plot 1  Plot 2  Plot 3  Variety 1 156,167,162,160,145,151,170,162169,168148,155 Variety 2 172,176,179,186,161,162,166,179160,176165,170 Variety 3 175,157,178,170,169,165,179,178172,174170,169\begin{array} { c | c c c } & \text { Plot 1 } & \text { Plot 2 } & \text { Plot 3 } \\\hline \text { Variety 1 } & 156,167 , & 162,160 , & 145,151 , \\& 170,162 & 169,168 & 148,155 \\\text { Variety 2 } & 172,176 , & 179,186 , & 161,162 , \\& 166,179 & 160,176 & 165,170 \\\text { Variety 3 } & 175,157 , & 178,170 , & 169,165 , \\& 179,178 & 172,174 & 170,169\end{array}

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H0: Variety has no effect on yield.
H1 ...

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ANOVA requires usage of the _______ distribution


A) Normal
B) t
C) Chi-square
D) F

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The following data contains task completion times, in minutes, categorized according to the gender of the machine operator and the machine used. The following data contains task completion times, in minutes, categorized according to the gender of the machine operator and the machine used.    The ANOVA results lead us to conclude that the completion times are not affected by an interaction between machine and gender, and the times are not affected by gender, but they are affected by the machine. Change the table entries so that there is an effect from the interaction between machine and gender. The ANOVA results lead us to conclude that the completion times are not affected by an interaction between machine and gender, and the times are not affected by gender, but they are affected by the machine. Change the table entries so that there is an effect from the interaction between machine and gender.

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The following table is one exa...

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Given below are the analysis of variance results from a Minitab display. to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. What can you conclude about the equality of the Population means?  Source DFSSMSFp Factor 313.5004.5005.170.011 Error 1613.9250.870 Total 1927.425\begin{array} { l | r | c | c | c | c } \text { Source } & \boldsymbol { D } \boldsymbol { F } & \boldsymbol { S S } & \boldsymbol { M S } & \boldsymbol { F } & \boldsymbol { p } \\\hline \text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\\hline \text { Error } & 16 & 13.925 & 0.870 & & \\\hline \text { Total } & 19 & 27.425 & & &\end{array}


A) Reject the null hypothesis since the P-value is greater than the significance level.
B) Accept the null hypothesis since the P-value is less than the significance level.
C) Accept the null hypothesis since the P-value is greater than the significance level.
D) Reject the null hypothesis since the P-value is less than the significance level.

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Draw an example of an F distribution and list the characteristics of the F distribution

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1) The F distribution is not symmetric...

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Select an appropriate null hypothesis for a one way analysis of variance test.


A) H0: μ\mu 1 = μ\mu 2 = μ\mu 3 = μ\mu 4
B) H0: μ\mu 1μ\mu 2μ\mu 3μ\mu 4
C) H0: p1 ≠ p2 ≠ p3 ≠ p4
D) H0 : σ1 = σ2 = σ3 = σ4

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Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance. At the 0.025 significance level, test the claim that the four brands have the same mean if the following sample results havebeen obtained.  Brand A  Brand B  Brand C  Brand D 1718212220182425212325272225262921262935293637\begin{array} { c c c c } \text { Brand A } & \text { Brand B } & \text { Brand C } & \text { Brand D } \\\hline 17 & 18 & 21 & 22 \\20 & 18 & 24 & 25 \\21 & 23 & 25 & 27 \\22 & 25 & 26 & 29 \\21 & 26 & 29 & 35 \\& & 29 & 36 \\& & & 37\end{array}

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blured image0 = blured image1 = blured image2

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Fill in the missing entries in the following partially completed one -way ANOVA table.  Source dfSSMS=SS/dfF-statistic  Treatment 311.16 Error 13.720.686\begin{array} { l | c | c | c | c } \text { Source } & d \boldsymbol { f } & \boldsymbol { S S } & \boldsymbol { M S } = \boldsymbol { S S } / d f & \boldsymbol { F } \text {-statistic } \\\hline \text { Treatment } & 3 & & & 11.16 \\\hline \text { Error } & & 13.72 & 0.686 &\end{array}


A)  Source dfSSMS=SS/dfF-statistic  Treatment 322.977.6611.16 Error 2013.720.686 Total 2336.69\begin{array} { l | c | c | c | c } \text { Source } & \boldsymbol { d f } & \boldsymbol { S S } & \boldsymbol { M S } = \boldsymbol { S S } / d \boldsymbol { f } & \boldsymbol { F } \text {-statistic } \\\hline \text { Treatment } & 3 & 22.97 & 7.66 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 36.69 & &\end{array}
B)  Source dfSSMS=SS/dfF-statistic  Treatment 32.557.6611.16 Error 2013.720.686 Total 2316.27\begin{array} { l | c | c | c | c } \text { Source } & d \boldsymbol { f } & \boldsymbol { S S } & \boldsymbol { M S } = \boldsymbol { S S } / d f & \boldsymbol { F } \text {-statistic } \\\hline \text { Treatment } & 3 & 2.55 & 7.66 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 16.27 & &\end{array}
C)  Source dfSSMS=SS/ddF-statistic  Treatment 348.8016.2711.16 Error 2013.720.686 Total 2362.52\begin{array} { l | c | c | c | c } \text { Source } & d f & \boldsymbol { S S } & \boldsymbol { M S } = \boldsymbol { S S } / d \boldsymbol { d } & \boldsymbol { F } \text {-statistic } \\\hline \text { Treatment } & 3 & 48.80 & 16.27 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 62.52 & &\end{array}
D)  Source dfSSMS=SS/dfF-statistic  Treatment 30.1840.06111.16 Error 2013.720.686 Total 2313.90\begin{array} { l | c | c | c | c } \text { Source } & \boldsymbol { d f } & \boldsymbol { S S } & \boldsymbol { M S } = \boldsymbol { S S } / \boldsymbol { d f } & \boldsymbol { F } \text {-statistic } \\\hline \text { Treatment } & 3 & 0.184 & 0.061 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 13.90 & &\end{array}

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Assume that the number of items produced is not affected by an interaction between employee and machine. Using a 0.05 significance level, test the claim that the machine has no effect on the number of items produced. Assume that the number of items produced is not affected by an interaction between employee and machine. Using a 0.05 significance level, test the claim that the machine has no effect on the number of items produced.   Using a 0.05 significance level, test the claim that the interaction between employee and machine has no effect on the number of items produced. Using a 0.05 significance level, test the claim that the interaction between employee and machine has no effect on the number of items produced.

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H0 : There is no interaction effect.
H...

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At the same time each day, a researcher records the temperature in each of three greenhouses. The table shows the temperatures in degrees Fahrenheit recorded for one week. At the same time each day, a researcher records the temperature in each of three greenhouses. The table shows the temperatures in degrees Fahrenheit recorded for one week.   i) Use a  0.05  significance level to test the claim that the average temperature is the same in each greenhouse. ii) How are the analysis of variance results affected if the same constant is added to every one of the original sample values? i) Use a 0.05 significance level to test the claim that the average temperature is the same in each greenhouse. ii) How are the analysis of variance results affected if the same constant is added to every one of the original sample values?

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i) Reject the claim that the a...

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Fill in the missing entries in the following partially completed one-way ANOVA table  Source dfSSMSF Treatment 313.89 Error 13.580.617 Total \begin{array} { l | l | l | l | l } \text { Source } & \boldsymbol { d f } & \boldsymbol { S S } & \boldsymbol { M S } & \boldsymbol { F } \\\hline \text { Treatment } & 3 & & & 13.89 \\\hline \text { Error } & & 13.58 & 0.617 & \\\hline \text { Total } & & & &\end{array}

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None...

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Given below are the analysis of variance results from a Minitab display. to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. Identify the P-value.  Source DFSSMSFp Factor 313.5004.5005.170.011 Error 1613.9250.870 Total 1927.425\begin{array} { l | r | c | l | c | c } \text { Source } & \boldsymbol { D F } & \boldsymbol { S S } & \boldsymbol { M S } & \boldsymbol { F } & \boldsymbol { p } \\\hline \text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\\hline \text { Error } & 16 & 13.925 & 0.870 & & \\\hline \text { Total } & 19 & 27.425 & & &\end{array}


A) 4.500
B) 5.17
C) 0.870
D) 0.011

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Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.  Exercise A  Exercise B  Exercise C2.55.84.38.84.96.27.31.15.89.87.88.15.11.27.9\begin{array} { c c c } \text { Exercise A } & \text { Exercise B } & \text { Exercise } \mathbf { C } \\\hline 2.5 & 5.8 & 4.3 \\8.8 & 4.9 & 6.2 \\7.3 & 1.1 & 5.8 \\9.8 & 7.8 & 8.1 \\5.1 & 1.2 & 7.9\end{array} At the 1% significance level, does it appear that a difference exists in the true mean weight loss produced by the three exercise programs?

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Test statistic: F = 1.491. Critical valu...

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Identify the value of the test statistic.  Source DFSSMSFp Factor 33010.001.60.264 Error 8506.25 Total 1180\begin{array} { l | r | c | c | c | c } \text { Source } & \boldsymbol { D } \boldsymbol { F } & \boldsymbol { S S } & \boldsymbol { M S } & \boldsymbol { F } & \boldsymbol { p } \\\hline \text { Factor } & 3 & 30 & 10.00 & 1.6 & 0.264 \\\hline \text { Error } & 8 & 50 & 6.25 & & \\\hline \text { Total } & 11 & 80 & & &\end{array}


A) 30
B) 1.6
C) 0.264
D) 10.00

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The data below represent the weight losses for people on three diets  Diet A  Diet B  Diet C 1.753.84.88.84.96.27.31.15.89.87.88.15.11.26.9\begin{array} { l c c } \text { Diet A } & \text { Diet B } & \text { Diet C } \\\hline 1.75 & 3.8 & 4.8 \\8.8 & 4.9 & 6.2 \\7.3 & 1.1 & 5.8 \\9.8 & 7.8 & 8.1 \\5.1 & 1.2 & 6.9\end{array} If we want to test the claim that the three size categories have the same means, why don't we use three separate hypothesis tests for μ1=μ2,μ2=μ3, and μ1=μ3?\mu _ { 1 } = \mu _ { 2 } , \mu _ { 2 } = \mu _ { 3 } , \text { and } \mu _ { 1 } = \mu _ { 3 } ?

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As we increase the number of i...

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