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9966 [12]
3 years ago
14

Caterpillar nitrogen assimilation and body mass. we explored the relationship between nitrogen assimilation and body mass (both

on log scales) for data on a sample of caterpillars in Caterpillars. The Instar variable in the data codes different stages (1 to 5) of caterpillar development.
a. Fit a model to predict log nitrogen assimilation (LogNassim) using log body mass (LogMass). Report the value of for this model.
b. Fit a model to predict LogNassim using appropriate indicators for the categories of Instar. Report the value of for this model and compare it to the model based on LogMass.
c. Give an interpretation (in context) for the first two coefficients of the fitted model in (b).
d. Fit a model to predict LogNassim using LogMass and appropriate indicators for Instar. Report the value of and compare it to the earlier models.
e. Is the LogMass variable really needed in the model of part (d). Does the linear trend appear to be better when the caterpillars are in a free growth period? (Again, we are not ready to fit more complicated models, but we are looking at the plot for linear trend in the two groups.)
Mathematics
1 answer:
Lady_Fox [76]3 years ago
3 0
A. Produce a scatterplot for predicting nitrogen assimilation (Nassim) based on Mass. Comment on any patterns.

b B.Produce a similar plot using the log (base 10) transformed variables, LogNassim versus Log Mass. Again, comment on any patterns.

C. Would you prefer the plot in part (a) or part (b) to predict the nitrogen assimilation of caterpillars with a linear model? Fit a linear regression model for the plot you chose and write down the prediction equation.
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Assume that females have pulse rates that are normally distributed with a mean of beats per minute and a standard deviation of b
snow_tiger [21]

Answer:

a) This is the p-value of Z when X = A subtracted by the p-value of Z when X = B.

b) P-value of Z when X = 76 subtracted by the p-value of X = 68.

c) Because the underlying distribution(pulse rates of females) is normal.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

Mean \mu, standard deviation \sigma.

standard deviation of beats per minute.

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is between A beats per minute and B beats per minute.

This is the p-value of Z when X = A subtracted by the p-value of Z when X = B.

b. If 4 adult females are selected, find the probability that they have pulse rates with a mean between 68 beats per minute and 76 beats per minute.

Sample of 4 means that we have n = 4, s = \frac{\sigma}{\sqrt{4}} = 0.5\sigma

The formula for the z-score is:

Z = \frac{X - \mu}{\sigma}

Z = \frac{X - \mu}{0.5\sigma}

Z = 2\frac{X - \mu}{\sigma}

This probability is the p-value of Z when X = 76 subtracted by the p-value of X = 68.

c. Why can the normal distribution be used in heartbeat even the sample side does not exceed 30?

Because the underlying distribution(pulse rates of females) is normal.

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