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Researchers Examined a New Treatment for Advanced Ovarian Cancer in a Mouse

Question 16

Multiple Choice

Researchers examined a new treatment for advanced ovarian cancer in a mouse model. They created a nanoparticle-based delivery system for a suicide-gene therapy to be delivered directly to the tumor cells. The mice were randomly assigned to have their tumor injected with either the gene -- nanoparticle combination, the gene alone, or some buffer solution (placebo) . The following table shows the tumor fold-increases after two weeks in a total of 29 mice. For reference, a fold-increase of 1 represents no change; a 2 represents a doubling in volume of the tumor.
 Buffer  Gene  Gene+Nano9.15.64.18.15.33.57.85.62.17.04.32.16.83.91.85.43.41.85.42.71.24.13.01.13.83.31.13.31.4\begin{array} { | r | r | r | } \hline \text { Buffer } & \text { Gene } & \text { Gene+Nano} \\\hline 9.1 & 5.6 & 4.1 \\8.1 & 5.3 & 3.5 \\7.8 & 5.6 & 2.1 \\7.0 & 4.3 & 2.1 \\6.8 & 3.9 & 1.8 \\5.4 & 3.4 & 1.8 \\5.4 & 2.7 & 1.2 \\4.1 & 3.0 & 1.1 \\3.8 & 3.3 & 1.1 \\3.3 & & 1.4 \\\hline\end{array} We want to test the null hypothesis that there is no difference in population mean tumor fold-increase for the three treatments. What are the correct summary statistics for these data? (Use technology.)


A)
 Buffer  Gene  Gene+Nano  mean 6.084.122.02 standard deviation 1.981.141.02\begin{array} { | r | c | c | c | } \hline& \text { Buffer } & \text { Gene } & \text { Gene+Nano } \\\hline \text { mean } & 6.08 & 4.12 & 2.02 \\\text { standard deviation } & 1.98 & 1.14 & 1.02 \\\hline\end{array}
B)
 Buffer  Gene  Gene+Nano  mean 5.084.122.02 standard deviation 3.921.291.04\begin{array} { | r | c | c | c | } \hline& \text { Buffer } & \text { Gene } & \text { Gene+Nano } \\\hline \text { mean } & 5.08 & 4.12 & 2.02 \\\text { standard deviation } & 3.92 & 1.29 & 1.04 \\\hline\end{array}
C)
 Buffer  Gene  Gene+Nano  mean 6.084.122.02 standard deviation 1.240.430.33\begin{array} { | r | c | c | c | } \hline& \text { Buffer } & \text { Gene } & \text { Gene+Nano } \\\hline \text { mean } & 6.08 & 4.12 & 2.02 \\\text { standard deviation } & 1.24 & 0.43 & 0.33 \\\hline\end{array}
D)  Buffer  Gene  Gene+Nano  mean 6.084.122.02 standard deviation 0.630.380.32\begin{array} { | r | c | c | c | } \hline& \text { Buffer } & \text { Gene } & \text { Gene+Nano } \\\hline \text { mean } & 6.08 & 4.12 & 2.02 \\\text { standard deviation } & 0.63 & 0.38 & 0.32 \\\hline\end{array}

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