Instructions: In one Excel le create one sheet and label it “Problem 1.” You sho
ID: 3049196 • Letter: I
Question
Instructions: In one Excel le create one sheet and label it “Problem 1.” You should clearly label each part of the problems with (a), (b), etc., and every computation should be labeled. For example, in the cell next to the mean of the data, you should have typed the word mean or average. Please make sure to do your computations using Excel. Do not do computations using Tables A and C or on a calculator and then type the answers into Excel. A random number generator is supposed to produce random numbers that are uniformly distributed on the interval from 0 to 1. If this is true, the numbers generated come from a population with µ = 0.5 and = 0.2887 . A command to generate 100 random numbers gives outcomes with mean ¯ x = 0.4365 . Assume that the population remains xed. We want to test H0 : µ = 0.5 and Ha : µ 6= 0.5 (4 points) Calculate the value of the z test statistic. (3 points) Is z statistically signicant at the 5% level ( = 0.05), show your work? (3 points) Is z statistically signicant at the 1% level ( = 0.01), show your work? (5 points) If 2.054 z 2.326, then between what two numbers does the P-value lie?
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 0.50
Alternative hypothesis: 0.50
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 0.02887
z = (x - ) / SE
z = - 2.199
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the z statistic less than -2.199 or greater than 2.199.
Thus, the P-value = 0.0278
Interpret results. Since the P-value (0.0278) is less than the significance level (0.05), we have to reject the null hypothesis.
Hence the result is statistically signicant at the 5% level.
Thus, the P-value = 0.0278
Interpret results. Since the P-value (0.0278) is greater than the significance level (0.01), we cannot reject the null hypothesis.
Hence the result is not statistically signicant at the 1% level.