A new study on the widely used antidepressant AD+ has been published. The study has information on the side effects of the drug. For each of the 24 randomly chosen users of AD+ who participated in the study, a score from 0 to 100 was assigned according to the severity and variety of side effects from the drug (a score of 0 meant no side effects). A similar assessment was made for 20 independently and randomly chosen users of the competitor antidepressant XAD24. The data are summarized in the following table: AD+ users XAD24 users =x146.1 =x246.8 =s21156.25 =s22457.96 (The first row gives the sample means and the second row gives the sample variances.) Assume that the populations of side effect scores for each of the drugs are approximately normally distributed. Can we conclude, at the 0.01 significance level, that the variance of scores for all AD+ users, σ21 , is different from the variance of scores for all XAD24 users, σ22 ? Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) The null hypothesis: H0: The alternative hypothesis: H1: The type of test statistic: (Choose one) Z t Chi square F The value of the test statistic: (Round to at least three decimal places.) The two critical values at the 0.01 level of significance: (Round to at least three decimal places.) and Can we conclude that the variance of scores for all AD+ users is different from the variance of scores for all XAD24 users? Yes No

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Description: A new study on the widely used antidepressant AD+ has been published. The study has information on the side effects of the drug. For each of the 24 randomly chosen users of AD+ who participated in the study, a score from 0 to 100 was assigned accordi

MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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A new study on the widely used antidepressant AD+ has been published. The study has information on the side effects of the drug. For each of the

randomly chosen users of AD+ who participated in the study, a score from

100

was assigned according to the severity and variety of side effects from the drug (a score of

meant no side effects). A similar assessment was made for

independently and randomly chosen users of the competitor antidepressant XAD24. The data are summarized in the following table:

AD+ users	XAD24 users
=x146.1	=x246.8
=s21156.25	=s22457.96

(The first row gives the sample means and the second row gives the sample variances.)

Assume that the populations of side effect scores for each of the drugs are approximately normally distributed. Can we conclude, at the

0.01

significance level, that the variance of scores for all AD+ users,

σ21

, is different from the variance of scores for all XAD24 users,

σ22

Perform a two-tailed test. Then fill in the table below.

Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)

The null hypothesis:	H0:
The alternative hypothesis:	H1:
The type of test statistic:	(Choose one) Z t Chi square F

The value of the test statistic: (Round to at least three decimal places.)
The two critical values at the 0.01 level of significance: (Round to at least three decimal places.)	and
Can we conclude that the variance of scores for all AD+ users is different from the variance of scores for all XAD24 users?	Yes	No

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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