【導(dǎo)讀】exampleoftime-seriesdata.exampleofcross-sectionaldata.equal.

　　

【正文】 statistic. z = (X bar μ0) / (σ /?√ n) Step 4: Specify the level of significance. You are willing to make a Type I error 5% of the time so the level of significance is . Step 5: State the decision rule regarding the hypothesis. The?≠ sign in the alternative hypothesis indicates that the test is twotailed with two rejection regions, one in each tail of the normal distribution curve. Because the total area of both rejection regions bined is (the significance level), the area of the rejection region in each tail is . The table value is +/ . This means that the null hypothesis should not be rejected if the puted z value lies between and +, and rejected if it lies outside of these critical values. Step 6: Collect the sample and calculate the test statistic. The value of X bar from the sample is . Since population standard deviation is given as , we calculate the z test statistic using σ as follows: z = ( ) / ( /?√ 49) = ( / ) = Step 7: Make a decision regarding the hypotheses. The calculated value of z = is called the calculated, puted, or observed test statistic. This zvalue indicates the location of the observed sample mean relative to the population mean ( sizeadjusted standard deviations to the left of the mean). Now, pare the calculated z to the critical z value. The calculated z value of is less than the critical value of , and it falls in the rejection region in the left tail. Reject H0. p: Identify the test statistic for a hypothesis test about the equality of two population means of two normally distributed populations based on independent samples. A test of differences in means requires using a test statistic chosen depending on two factors: 1) whether the population variances are equal, and 2) whether the population variances are known. This test can be used to test the null hypothesis: H0: μ1 μ2 = 0 vs. the alternative of HA: μ1 μ2?≠ 0 The test when population means are normally distributed and the population variances are unknown but assumed equal uses a pooled variance. Use the tdistributed test statistic: t = {(X bar1 X bar2) (μ1 μ2)} / {(sp2 / n1) + (sp2 / n2)}1/2 The test when the population means are normally distributed and population variances unknown and cannot be assumed to be equal uses the tdistributed test statistic that uses both samples39。 variances: t = {(X bar1 X bar2) (μ1 μ2)} / {(s12 / n1) + (s22 / n2)}1/2 q: Formulate a null and an alternative hypothesis about the equality of two population means (normally distributed populations, independent samples) and determine whether the null hypothesis is rejected at a given level of significance. Sue Smith is investigating whether the announcement period abnormal returns to acquiring firms differ for horizontal and vertical mergers. She estimated the abnormal announcement period returns for a sample of acquiring firms associated with horizontal mergers and a sample of acquiring firms involved in vertical mergers. She estimated the abnormal announcement period returns for a sample of acquiring firms associated with horizontal mergers and a sample of acquiring firms involved in vertical mergers. Abnormal returns for horizontal mergers: mean = 1%, standard deviation = 1%, sample size = 64. Abnormal returns for vertical mergers: mean = %, standard deviation %, sample size = 81. Assume that the population means are normally distributed, and that the population variances are equal. Is there a statistically significant difference in the announcement period abnormal returns for these two types of mergers? Step 1: State the hypothesis. H0: 181。1 181。2 = 0 HA: 181。1 181。2 ≠ 0 Where 181。1 is the mean of the abnormal returns for the horizontal mergers and 181。2 is the mean of the abnormal returns for the vertical mergers. Step 2: Select the appropriate test statistic. Use the test statistic formula that assumes equal variances (from LOS ). Step 3: Specify the level of significance. There are n1 + n2 2 degrees of freedom. Therefore, there are 64 + 81 2 = 143 degrees of freedom. We will use the mon significance level of 5%. Step 4: State the decision rule regarding the hypothesis. Using the tdistribution table and the closest degrees of freedom, the critical value for a 5% level of significance is . Step 5: Collect the sample and calculate the sample statistics. Inputting our data into the formula: sp2 = (63)() + (80)() / 143 = t = / = Step 6: Make a decision regarding the hypothesis. Because the calculated test statistic falls to the left of the lowest critical value, we reject the null hypothesis. We conclude that the announcement period abnormal returns are different for horizontal and vertical managers. r: Identify the test statistic for a hypothesis test about the mean difference for two normal distributions (paired parisons test). Frequently, we are interested in the difference of paired observations. If questions about the differences can be expressed as a hypothesis, we can test the validity of the hypothesis. When paired observations are pared, the test bees a test of paired differences. The hypothesis bees: H0: 181。d = 181。d0 HA: 181。d ≠ 181。d0 Where 181。d is the population mean of differences and 181。d0 is the hypothesized difference in means (often zero). the alternative may be onesided: HA: 181。d 181。d0 HA: 181。d 181。d0 For paired differences, the test statistic is: t = {(d bar 181。d0) / sd bar}. s: Formulate a null and an alternative hypothesis about the mean difference of two normal populations (paired parisons test) and determine whether the null hypothesis is rejected at a given level of significance. Joe Andrews is examining changes in estimated betas for the mon stock of panies in the telemunications industry before and after deregulation. Joe believes that the betas may decline because of deregulation, be