17. We construct a contingency table to compare multiple population p tions. Con

Question

17. We construct a contingency table to compare multiple population p tions. Contingency tables also allow to test for (a) independence of populations (b) errors in calculation (c) differences in population mean (d) variance in population proportion 18, what is the p-value for : = 1.95? (a) 0.9744 (b) 0.0256 (c) 0.0314 (d) 0.3594 19. Suppose we are testing hypotheses Ho : --0 and Hi : -P? We calculate the test statistic and find a corresponding p-value of 0.02 What evidence have we found? (a) no evidence supporting H (b) weak evidence supporting H (c) strong evidence supporting H (d) overwhelming evidence supporting H 20. The ANOVA analysis requires our samples to be (a distributed (b) approximately normal-distributed (c) uniformly distributed (d) exponentially distributed

Explanation / Answer

Answer to all question is as follows:

1. b is the unbiased estimator of mu1-mu2

2. The test statistic for mu1-mu2 when the variances are equal an independent sample t-test
So, b is right

3. The subscript given are alpha and df = .01,30 we refer to the 2 tailed distribution.
we have a t value from the t tables as 2.750. C is right

4. We will not reject null hypothesis as our tstatistic is not within /beyond the critical value
or tstat<tcritical (1.4<2)
Hence, d) we do not accept Null hypothesis

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