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In an opinion survey, a random sample of 1000 adults from the U.S.A and 1000 adults from Germany were asked whether they supported the death penalty. 590 American adults and 560 German adults indicated that they supported the death penalty. The researcher wants to know if there is sufficient evidence to conclude that the proportion of adults who support the death penalty is higher in the U.S.A. than in Germany. What is the decision at a =.05 and 0.10
A machine is made up of two independent components placed in series. The lifetime of each component is uniformly distributed over the interval [0,1].
Research on new juvenile delinquents revealed that 38% of them committed another crime. What is the probability that 40 or fewer of the delinquents will commit another crime?
you are given the accompanying response data on concentration of a chemical as a function of time. the six sets of
Under what conditions is a grouped frequency distribution more useful as a means of describing a set of scores than a simple frequency distribution?
1. classify the following variable as nominal ordinal discrete or continuous and state why it is such a variable
Let (X, S) be a measurable space and En measurable sets, not necessarily disjoint, whose union is X. Suppose that for each n, fn is a measurable real-valued function on En. Suppose that for any x ∈ Em ∩ En for any m and n, fm (x ) = fn (x ). Le..
A sample of 140 golfers showed that their average score on a particular golf course was 93.38 with a standard deviation of 4.36. Answer each of the following (show all work and state the final answer to at least two decimal places.):
find the 98 confidence interval for the difference between two means based on this information about two samples.
heights of trees in a forest are normally distributed. the mean and standard deviation of a random sample of size 20
a. What ids the population? b. What is the sample? c. Is this a judgment sample of a probability sample?
compute the pearson correlation for the following set of scoresx 2 3 4 2 4y 2 7 6 4 6can you please show your
(b) of Exercise 25 to calculate -2 · ln λ and test the null hypothesis = = = at the 0.05 level of signifi- cance. Explain why the number of degrees of freedom for this approximate chi-square test is 3.
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