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tatiyna
3 years ago
11

A forester wishes to predict the volume (in cubic feet) of usable lumber in a certain species of tree using the height (in feet)

and diameter (in inches) of the trees. The height and diameter of 31 trees of a certain species were measured, the trees were cut down, and the volume of usable lumber was determined. The response and explanatory variable(s) in this study are:
Mathematics
1 answer:
alex41 [277]3 years ago
7 0

Answer: The response variable = Volume

and explanatory variable(s) in this study are: Height and Diameter.

Step-by-step explanation:

Important definitions:

  • The explanatory variable is a type of independent variable manipulated by the experimenter for the ongoing experimental study.
  • The response variable is a type of dependent variable which depends on the explanatory values provided by the experimenter.

Given: A forester wishes to predict the volume (in cubic feet) of usable lumber in a certain species of tree using the height (in feet) and diameter (in inches) of the trees.

Here, Volume directly depends upon the height and diameter of the tree.

Also, forester ( Experimenter) wants to predict volume by cutting tree that means it is an response variable and it depends on height and diameter , so height and diameter are explanatory variable.

Hence, The response variable = Volume

and explanatory variable(s) in this study are: Height and Diameter.

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A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

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                               P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average time = 23.92 minutes

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<em> Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

<u>So, a 98% confidence interval for the population mean, </u>\mu<u> is; </u>

P(-2.426 < t_3_9 < 2.426) = 0.98  {As the critical value of z at 1%  level

                                               of significance are -2.426 & 2.426}  

P(-2.426 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.426) = 0.98

P( -2.426 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

P( \bar X-2.426 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

<u>98% confidence interval for</u> \mu = [ \bar X-2.426 \times {\frac{s}{\sqrt{n} } } , \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ]

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Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

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