Reference no: EM13916551
normally distributed and the population standard deviation is known. She uses a Z test to test the null hypothesis that the mean tensile strength is 800 pounds per square inch. The calculated Z test statistic is a positive value that leads to a p-value of .009 for the test. If the significance level (a) is .01, the null hypothesis would be rejected.
2. You cannot make a Type II error when the null hypothesis is true.
3. The power of a statistical test is the probability of not rejecting the null hypothesis when it is true.
4. The level of significance indicates the probability of rejecting a false null hypothesis.
5. When conducting a hypothesis test about a single mean, other relevant factors held constant, increasing the level of significance from .05 to .10 will decrease the probability of a Type II error.
Chapter10
6. In an experiment involving matched pairs, a sample of 14 pairs of observations is collected. The degree of freedom for the t statistic is 12.
7. In testing the difference between two means from two independent populations, the sample sizes do not have to be equal to be able to use the Z statistic.
8. In testing the difference between the means of two independent populations, if neither population is normally distributed, then the sampling distribution of the difference in means will be approximately normal provided that the sum of the sample sizes obtained from the two populations is at least 30.
9. If the limits of the confidence interval of the difference between the means of two normally distributed populations were 0.5 and 2.5 at the 95% confidence level, then we can conclude that we are 95% certain that there is no significant difference between the two population means.
10. When comparing two population means based on independent random samples, the pooled estimate of the variance is used if the population standard deviations are not known and assumed unequal.
Chapter 12
11. The chi-square distribution is a continuous probability distribution that is always skewed to the right.
12. In a contingency table, when all the expected frequencies equal the observed frequencies the calculated c2 statistic equals 1.
13. In a contingency table, if all of the expected frequencies equal the observed frequencies, then we can conclude that there is a perfect dependence between rows and columns.
14. In performing a chi-square test of independence, as the difference between the respective observed and expected frequencies decrease, the probability of concluding that the row variable is independent of the column variable increases.
15. When we carry out a chi-square test of independence, the expected frequencies are based on the Null hypothesis.
Multiple Choices (Two points each)
Chapter 9
1. If a null hypothesis is rejected at a significance level of .05, it will ______ be rejected at a significance level of .01
A. Always
B. Sometimes
C. Never
2. If a null hypothesis is not rejected at a significance level of .10, it will ______ be rejected at a significance level of .05
A. Always
B. Sometimes
C. Never
3. When carrying out a large sample test of H0: m =10 vs. Ha: m ¹ 10 by using a p-value, we reject H0 at level of significance a when the p-value is:
A. Greater than a/2
B. Greater than a
C. Less than a
D. Less than a/2
E. Less than Za
4. If you live in California, the decision to buy earthquake insurance is an important one. A survey revealed that only 119 of 340 randomly selected residences in one California County were protected by earthquake insurance.
Calculate the appropriate test statistic to test the hypotheses that 40% or more buy the insurance. (You are dealing with proportions which are small numbers. Therefore keep at least four decimal places in your intermediate steps and do the rounding to the two decimal places at the end only).
A. 1.88
B. 1.93
C. -1.88