If the underlying theory does not suggest a particular direction for the value of an estimated variable, it is appropriate to use a:
Answer(s): C
If there is no suggested directionality to the underlying variable, then you are interested in finding the probability under the null hypothesis that you will find a value beyond a given distance from the null value. For this, you need to use a two-tailed test.
In a binomial probability distribution,
Answer(s): D
A binomial distribution with "N trials" arises when the same experiment is repeated N times. Each of the trails is known as a "Bernoulli" trial. It is the Bernoulli trial which can only take two distinct values, "success" and "failure." Each outcome of a binomial distribution with N trials is the string of N values generated by the individual Bernoulli trials. Clearly, the binomial distribution with N trials can take 2^N values, even though each individual trial can have only two outcomes. It is important that you understand the difference between the binomial distribution and the underlying Bernoulli trials themselves.
Which of the following are notations for the covariance between X and Y?
I and II are both notations for covariance between X and Y.
A sample of size 225 is drawn from a population. The sample mean equals 876 and the variance of the sample equals 5,924. The 85% confidence interval for the population mean is given by:
If z is the z-value corresponding to the specified confidence level, the sample mean is M and the standard deviation is D in a sample size N, the confidence interval is specified as [M - z*D/sqrt(N), M + z*D/sqrt(N)]. In the present case, for the 85% confidence interval, the normal probability table gives z = 1.44. The sample standard deviation equals sqrt(5924) = 76.97. Therefore, the confidence interval equals [876 - 1.44*76.97/sqrt (225), 876 + 1.44*76.97/sqrt(225)] = [868.6, 883.4]
The seasonal output of a new experimental strain of pepper plants was carefully weighed. The mean weight per plant is 15.0 pounds, and the standard deviation of the normally distributed weights is 1.75 pounds. Of the 200 plants in the experiment, how many produced peppers weighing between 13 and 16 pounds?
z = (x-u)/sigma. z1 = 13 - 15/1.75 = -1.1429 and z2 = 16 - 15/1.75 = 0.5714. The respective areas for those z- values are 0.3729 and 0.2157. Since they are on opposite sides of the mean, we add them up to find the area in between which is 0.5886. Therefore, 0.5886*200 = 118.
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