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kobusy [5.1K]
3 years ago
14

Measuring the height of a tree is usually more difficult than measuring the diameter of the tree. Therefore, many researchers us

e regression models to predict the height of a tree from its diameter measured at 4 feet 6 inches from the ground. The following computer output shows the results of a linear regression based on the heights, in feet, and the diameters in inches, recorded from 31 felled trees.
Estimate Std Error t-value Pr(>|t|)
Intercept 62.031 4.383 14.15 0.0000
Diameter 1.054 0.322 0.0028 3.27
Which of the following is a 95 percent confidence interval for the slope of the population regression line?
(A) (0.001.2.107).
(B) (0.396, 1.712.
(C) (0.423,1.685).
(D) (0.732. 1.376).
(E) (53.07.70.99).
Mathematics
1 answer:
LenKa [72]3 years ago
8 0

Answer:

(B) (0.396, 1.712)

Step-by-step explanation:

From the information given;

Confidence Interval = 0.95

Significance Level = 1 - 0.95 = 0.05

The confidence interval for regression coefficient beta (whereby in this case it is the coefficient of the diameter) is expressed by:

= ( Beta ^- \ t_{n-2}, \alpha/2 ( S.E \ of \ \^ \ \ ), Beta^+ \ t_{n-2}, \alpha/2 ( S.E \ of \ \^ \ \ ))

From the regression coefficient, the estimated value of beta^  = 1.054

= (1.054 - t_{29,0.025} *(0.322), 1.054 + t_{29,0.025}*(0.322))

t_{229,0.025} = 2.0452\\\\ = \mathbf{(0.396, 1.712)}

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Data from the Centers for Disease Control and Prevention indicate that weights of American adults in 2005 had a mean of 167 poun
sesenic [268]

Answer:

92.65% probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

For sums the theorem can also be used, with mean n*\mu and standard deviation s = \sqrt{n}*\sigma

In this problem, we have that:

n = 47, \mu = 47*167 = 7849, s = \sqrt{47}*35 = 240

Use this information to estimate the probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

This is 1 subtracted by the pvalue of Z when X = 7500. So

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

By the Central Limit Theorem

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

Z = \frac{7500 - 7849}{240}

Z = -1.45

Z = -1.45 has a pvalue of 0.0735

1 - 0.0735 = 0.9265

92.65% probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

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Step-by-step explanation:

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