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2 years ago

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A company performs linear regressions to compare data sets of two similar products. If the residuals for brand A form an increas

ing curve, and the residuals for brand B form a U-shaped pattern, what can be concluded?
A. Both data sets are probably linear.
B. Neither data set is likely to be linear.
C. Brand B’s data are probably linear, while brand A’s data are probably not.
D. Brand A’s data are probably linear, while brand B’s data are probably not.

2 answers:

liberstina [14]2 years ago

4 0

D, IS THE FINAL ANSWER :)

Bogdan [553]2 years ago

3 0

Leaning towards D. -not too sure though

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Then, the expected mean and the standard deviation of the sampling distribution will be :_

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1 year ago

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, and then state the concl

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a) Failed to reject the null hypothesis (P-value=0.09).

b) The 95% CI for the difference in proportions is:

$-0.0599\leq\pi_1-\pi_2\leq0.0124$

Step-by-step explanation:

a) We have to perform a hypothesis test for the difference of proportions.

The null and alternative hypothesis are:

$H_0: \pi_1\geq\pi_2\\\\H_1: \pi_1$

The significance level is 0.05.

The proportion of the passenger cars owners is:

$p_1=\frac{239}{2142} =0.1116$

The proportion of commercial truck owners is:

$p_2=\frac{54}{399}=0.1353$

The weigthed average p is

$p=\frac{n_1p_1+n_2p_2}{n_1+n_2}=\frac{239+54}{2142+399}=0.1153$

The estimated standard deviation is

$s=\sqrt{\frac{p(1-p)}{n_1}+\frac{p(1-p)}{n_2}} =\sqrt{\frac{0.1153(1-0.1153)}{2142}+\frac{0.1153(1-0.1153)}{399}} =0.0174$

We can calculate the z-value as:

$z=\frac{\Delta p}{s}=\frac{0.1116-0.1353}{0.0174}=-1.362$

The P-value for z=-1.362 is P=0.0866.

The P-value (0.09) is greater than the significance level (0.05), so it failed to reject the null hypothesis. There is no enough evidence to prove that commercial trucks owners violate laws requiring front license plates at a higher rate than owners of passenger cars.

b) We can construct a 95% CI, according to the significance level of 0.05.

The z-value for this CI is 1.96.

We have to recalculate the standard deviation:

$\sigma=\sqrt{\frac{p_1(1-p_1)}{n_1} +\frac{p_2(1-p_2)}{n_2}} =\sqrt{\frac{0.1116(1-0.1116)}{2142} +\frac{0.1353(1-0.1353)}{399}} =0.0184$

The lower limit is then:

$LL=(p_1-p_2)-z*\sigma=(0.1116-0.1353)-1.96*0.0184=-0.0238-0.0361\\\\LL=-0.0599$

The upper limit is:

$UL=(p_1-p_2)+z*\sigma=(0.1116-0.1353)+1.96*0.0184=-0.0238+0.0361\\\\UL=0.0124$

The 95% CI for the difference in proportions is:

$-0.0599\leq\pi_1-\pi_2\leq0.0124$

In this case, we can conclude that the difference between the proportions, with 95% confidence, can still be equal or greater than zero, meaning that it is possible passenger car owners violate laws more than truck owners.

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2 years ago