H0: µ1 – µ2 = 0
H1: µ1 – µ2 ≠ 0
or
H0: µ1 = µ2 (There is no difference)
H1: µ1 ≠ µ2 (There is a difference)
Statıstıcs II (ENG) — Ünite 4 Soru-Cevap
Statıstıcs II (ENG) (IST210U) soru-cevapları.
For two sided (two-tailed) test, express the null (H0) and alternative (H1) hypothesis.
For the right sided (upper tailed) test, express null (H0) and alternative (H1) hypothesis.
H0: µ1 – µ2 = 0
H1: µ1 – µ2 > 0
or
H0: µ1 = µ2
H1: µ1 > µ2
Express the test statistic for two sample tests of means when their population variances are known, in terms of difference of the sample means, population variances, and the sizes of the sample of the two independent populations.
Express the rejection regions for the following cases:
If the alternative hypothesis is in the form of H1 : µ1 ≠ µ2
If the alternative hypothesis is in the form of H1 : µ1 < µ2
If the alternative hypothesis is in the form of H1 : µ1 > µ2
If the alternative hypothesis is in the form of H1 : µ1 ≠ µ2 then rejection region is
z > zα/2 or z < – zα/2
If the alternative hypothesis is in the form of H1 : µ1 < µ2 then rejection region is
z < – zα
If the alternative hypothesis is in the form of H1 : µ1 > µ2 then rejection region is
z > zα
A statistics lecturer teaches the same statistics course to two different groups. One group is from the department of economics and the other group is from the department of business. The lecturer has been teaching these two departments for a long time. He/she has a pretty good idea about the success of the students on statistics course. The final grades of the students for both departments follow a normal distribution. The lecturer has calculated that the variances of the final exam grades up to this term for the business and economics students is 12 and 14 respectively. This term, 120 students from department of business and 140 students from department economics have enrolled in the statistics course. Clearly there are two populations here, students from Business and Economics Departments. Let’s use the enrolled students from these departments as samples. At the end of term, final grade means of for the business department and the economics department students are 62 and 65, out of 100, respectively. Is there a difference of the mean grades of students taking the course from these two different departments, choose the significance level as 5%?
We can apply the 5-step procedure of the hypothesis testing to solve this problem as follows:
Step 1: In this question, the, parameter of interest is the difference of the mean final grades of the students from different groups. Therefore, the null hypothesis (H0
) is “there is no difference between final grades of the students from different departments”, and the alternative hypothesis (H1) is “there is a difference between final grades of the students from different departments”. Then, these hypotheses can be expressed as follows:
H0: µ1 – µ2 = 0 (There is no difference between final grades)
H1: µ1 – µ2 ≠ 0 (There is a difference between final grades)
Step 2: It’s indicated in the problem that the significance level is α = 0.05 for this problem. Note that the significance level α = 0.05 also specifies the type I error probability of rejecting the null hypothesis (H0) when it’s actually true.
Step 3: The independent random sample sizes are n1= 120 students and n2= 140 students. Also, the sample mean grades are x1= 62 and x2 = 65. Then, the test statistic can be obtained as follows:
Step 4: The decision rule for two-sided test at the level of significance α = 0.05 is shown in the Figure 4.1, and we utilize the standard normal distribution, and standard normal distribution table values are given in Table 4.1. According to Table 3.1, we reject the null hypothesis (H0 ) if the calculated value of z is less than – 1.96 and greater than 1.96. On the other hand, we fail to reject the null hypothesis (H0 ) if z lies between – 1.96 and 1.96. H1 : µ1 ≠ µ2 , then rejection region is z > zα/2 or z < – zα/2. Explicitly the rejection region is z > 1.96 or z < – 1.96
Step 5: Conclusion: Since the test statistic value is ztable = 1.96 < z = 6.71, we fail to reject the null hypothesis H0: µ1 = µ2 at the level of significance α = 0.05.
As a result, we can conclude that there is no difference between final grades of the students from different departments at the level of significance α = 0.05.
A random sample of 40 students is selected from university (A) and their particular subject mean test score is 65 with a standard deviation 2.7. Another random sample of 60 students is selected from university (B) and their mean test score is 61 with a standard deviation 4.5. At the 1% significance level, can we conclude that university (A) mean test score is better than the university (B).
We can apply the 5-step procedure of the hypothesis testing to solve this problem.
Step 1: In this question, the parameter of interest is the difference of the mean test scores. Therefore, the null hypothesis (H0) is “there is no difference between the test scores of the students”, and the alternative hypothesis (H1) is “university (A) students' test scores are better than the university B students' test scores. Then, these hypotheses can be expressed as follows:
H0: µA = µB
H1: µA > µB
Step 2: It’s indicated in the problem that the significance level is α = 0.01 for this problem.
Step 3: The independent random sample sizes are nA = 40 students and nB = 65 students. Also, the university (A) mean test score is xA = 65 and standard deviation sA = 2.7, and the university (B) mean test score is xB = 61 and standard deviation sB= 4.5.
Then, the test statistic can be obtained as follows:
Step 4: The decision rule for one sided test, right sided (upper tailed) test at the level of significance α = 0.01 is shown in the Figure 4.2, and we utilize the standard normal distribution and standard normal table to obtain the critical value for α = 0.01 from Table 4.1 (s.79).
According to Table 4.1, we reject the null hypothesis (H0) if the calculated value of z is greater than 2.33. H1: µA > µB , then rejection region is z > zα. Explicitly, the rejection region is z > 2.33
Step 5: Conclusion: Since the test statistic zcal = 5.69 > z = 2.33, we reject the null hypothesis H0: µA = µB at the level of significance α = 0.01. According to this result at the α = 0.01 significance level, we can conclude that university (A) mean test score is better than the university (B) mean test score.
Explain the use of t test instead of z test in terms of sample size.
z test statistic can be used for difference between two means if sample sizes are greater than or equal to 30. In many real-life circumstances, it’s not possible or practical to attain the sample size of 30 or more due to the time, research budget and other relevant constraints. Also, usually the population standard deviations are unknown. Under these circumstances, t test is used to test the difference between means. At this time also, the assumption regarding the population distribution of the variables should be normal or approximately normal, and two samples are independently selected from these populations.
Under the assumption of unequal variances for testing the difference between two
means, formulate the test statistic t.
If the population variances are assumed to be equal for testing the difference between two means, formulate the test statistic t.
The students are prepared for the English proficiency exam in two small groups. The sizes of the groups are 8 and 10 students, respectively. The average score of the first group is 80 from the test and the average score of the second group is 72. Also, standard deviation of the test scores are s1= 9 and s2= 5, respectively. Test the claim that the first group’s (the size of 8) proficiency exam scores are greater than the other group (the size of 8) with a α = 0.10 significance level. It’s assumed that the populations are normally distributed, and the population variances are equal.
We can apply the 5-step procedure of the hypothesis testing to solve this problem.
Step 1: In this question, the parameter of interest is the difference of the mean English proficiency exam test scores of two groups of size n1= 8 and n2= 10. Since the claim in the question is the first group’s proficiency exam scores are greater than the second group, the alternative hypothesis is right tailed (right sided). Then hypothesis can be expressed as follows:
H0: µ1 = µ2 or H0: µ1 – µ2 = 0
H1: µ1 > µ2 or H1: µ1 – µ2 > 0
Step 2: It’s indicated in the problem that the significance level is α = 0.10 for this problem.
Step 3: In this problem, it’s assumed that the population variances are equal and the degrees of freedom are d. f. = n1+ n2– 2 = 8 + 10 – 2 = 16
The test statistic can be obtained as follows:
Step 4: The decision rule for one sided test at the level of significance α = 0.10 is shown in the Figure 4.4 (s. 84), and we utilize the t table to obtain the critical value for α = 0.10 for d. f. = 16
Since the hypothesis test that we considered is right sided (one tailed) test with d. f. = 16 and α = 0.1, the critical value is t0 = 1.337. Consequently, the rejection region is t > 1.337, as shown in Figure 4.4.
Step 5: Conclusion: Since the test statistic t = 2.397 > t0 = 1.337, we reject the null hypothesis H0: µ1 = µ2 at the level of significance α = 0.10. According to this result at the α = 0.10 significance level, we can conclude that the mean proficiency exam score of the first group is significantly higher than the second group.
Covering the hypothesis test of the difference between two population proportions, and these parameters are depicted with p1 and p2, a sample proportion from each population is utilized to realize the z-test for the difference between two population proportions. The basic assumption of the test considered here is that there is no difference between the population proportions. Then, define the hypotheses for difference between two population proportions for two sided test and express null and alternative hypothesis.
For two sided (two tailed) test, the null hypothesis (H0) and the alternative hypothesis (H1) are in the following form,
H0: p1 = p2
H1: p1 ≠ p2
Covering the hypothesis test of the difference between two population proportions, and these parameters are depicted with p1 and p2, a sample proportion from each population is utilized to realize the z-test for the difference between two population proportions. The basic assumption of the test considered here is that there is no difference between the population proportions. Then, define the hypotheses for difference between two population proportions for right sided test and express null and alternative hypothesis.
For the right sided (upper tailed) test, the null hypothesis (H0) and the alternative hypothesis (H1) are in the following form,
H0: p1 = p2
H1: p1 > p2
Covering the hypothesis test of the difference between two population proportions, and these parameters are depicted with p1 and p2, a sample proportion from each population is utilized to realize the z-test for the difference between two population proportions. The basic assumption of the test considered here is that there is no difference between the population proportions. Then, define the hypotheses for difference between two population proportions for left sided test and express null and alternative hypothesis.
For the left sided (lower tailed) test, the null hypothesis (H0) and the alternative hypothesis (H1) are in the following form,
H0: p1 = p2
H1: p1 < p2
Explain the application of z test for the difference between two population proportions.
To apply the z test for the difference between two population proportions; the samples must be independent and selected randomly. Since standard normal distribution can be used to test statistical hypothesis for proportions; then, the fallowing condition for proportions and sample sizes must be satisfied; n1p1≥ 5, n1q1≥ 5, n2p2≥ 5, n2q2≥ 5 where q1= 1 – p1 and q2 = 1 – p2.
Then, z-test is used to test the difference between two proportions p1 and p2 and the test statistic as follows,
When there are more than two means define null and alternative hypothesis.
The null
hypothesis and alternative hypothesis for one-way analysis of variance test can be defined as follows,
H0: µ1 = µ2 = µ3 = ..... = µk (all population means under consideration are equal)
H1: At least one of population means is different µ1, µ2, µ3, ..... , µk.
Rejecting the null hypothesis, explain the practice to find the difference for more than two means.
rejecting the null hypothesis in an ANOVA test means that at least one of the means of the
population under consideration is different from the other population's means. To determine which of the population mean(s) are different we need additional statistical tests. F test is used if we consider more than two means of population. By the help of the F test, all the population means are compared simultaneously. Otherwise, if you consider comparing two means at a time as we considered in the previous sections, the number of tests increases enormously as the number of populations and their means increase. For example, to compare three population's means at a time, we need three t-tests. In a similar manner, to compare five population's means at a time, we need 10 t-tests.
What are the requirements to utilize one-way ANOVA test ?
To utilize the one-way ANOVA test, the samples must be independent and randomly selected from a normal or approximately normal population. Also, each population variance under consideration must be equal.
Explain the one-way ANOVA test.
For the one-way ANOVA test, the test statistic is the ratio of two variances: namely the variance between samples and the variance within samples. These variances constitute the estimates of the population variance. The variance between samples are also called as between group variance, and the variance within samples are also called as within group variance. The variance between samples measures the differences associated with the treatment which is given to each sample and abbreviated as MSB . On the other hand, the variance within samples measures the differences associated with observations within the same sample abbreviated as MSW . The test statistic for a one-way ANOVA test is the ratio of two variances; namely the variance between samples (MSB) and variance within samples (MSW).
When comparing more than two means, explain the rejection of null hypothesis.
The F (one-way ANOVA) test to compare more than two population means is always a right tailed test. Therefore, H0 will be rejected if the test statistic F is greater than the critical value. From the F test statistic, it’s clear that the value of the test statistic is close to the 1 if MSB close MSW , means that there is little or no difference between the means. Then, the F test statistic value nearby to 1 suggests that we fail to reject the null hypothesis (H0).
If F test statistic value is greater than 1 we suggest that the null hypothesis should be rejected. This situation occurs when one population mean is significantly different than the other population's means, and this state is revealed if MSB is greater than MSW .
When comparing more than two means, how is the sum of squares within the groups and between groups can be calculated?
