I'll give you a few hints.
The null hypothesis ALWAYS uses an equals sign. It could be greater than or equal to or less than or equal to or just equals some value.
An alternate hypothesis can be greater than, less than, or does not equal some value. If no direction is specified (does not equal), then the test is two-tailed. Otherwise, it is one-tailed.
For #2, try a one-sample z-test. Use a z-table to determine your critical value for a two-tailed test (the alternate hypothesis will not show a specific direction since the problem just wants to know if there is a change and that change could be in either direction). Compare the critical value to the test statistic. If the test statistic exceeds the critical value in either direction from the table, reject the null and conclude a change. If the test statistic does not exceed the critical value in either direction from the table, do not reject the null.
For #3, try a one-sample t-test since the sample size is rather small. Use a t-table to find the critical value for a two-tailed (nondirectional) test. Go through the same process as #2.
It looks like you are asked to use SPSS for #4. Try a z-test.
For #5, the alpha level directly affects the power of a test. Sample size also affects power. The test has greater power at .05 than .01.
Hi I need help with Statistics. Can anyone help me to understand it better? I am in fear that if I do not figure this out soon I will fail my class. I am needing help with the chapters 12,and 13. in the book called Statistical Reasoning in the Behavioral Sciences, 6th edition.
1. State whether each of the items in the list could be an alternative hypothesis and if so whether it would be a one tailed test or a two tailed test.
x- >345
µx≠12
x-=67
µx<89
x-<123
µx=45
µx>678
x-≠910
2. Two years ago, a complete survey of all male students in a large university indicated that the mean number of males that smoked per day was 8.3 with a standard deviation of 3.7. the director of the school wanted to see if the habits changed over the past two years. He got these results from the students: x- =7.7 n=108.
set up Ho and Ha
Perform the statical test (α =.05)
draw a conclusion.
3. the PTA at central H.S. is concerned that the students who graduate from the school do not score as well on the math test, on average, as do students from other school district. to address their concerns the principle randomly selected 15 students from the graduating class and administers the math teat used in that school district. the mean of the scores is 98. the scores of the 15 students from central high school are as follows: 105,98, 101, 110,96, 103, 104, 101, 98, 105, 112, 95, 105, 100, 108.
state formally the hypotheses necessary to conduct a nondirectional test
complete the test at the .05 level of significance and state your conclusion.
4.an industral/orginizational psychologist used a new training program designed to improve basic work shills. the traing program is typically judged as effective if average scores on a basic work skills test exceed 80. the psychologist obtained a random sample of participants who recently completed the training program, then administered to them a work skills test. their scores on the trest are givin in data 12A.
Data 12A
state the null and alt. hypothesis best suited to the nature of the psychologest inquiry
using IBM SPSS test the null hypothesis at the .01 level of significance and state your conclusion
5. with regard to Data 13A...
if α = .05 and µtrue=83.0 what is β
what is the power of the test?
if α=.01 and µtrue = 83.0 what is β
what is the power of the test?
Data 12A:
83 78 85 85 82 74 79 78 88 80 85 81 85 88 84 86 87 87 82 87 87 87 82 91 75 85 82 92 87 83 84 94 84 77 86 89 83 89 90 82 94 97 84 78 90 82 86 80 88 89 82 86 88 82 75 77 94 79 88 89
Data 13A:
Ho:µx= 80.0 Ha:µx≠80.0 σx=20.0 n=100
2 answers
Get an answer from tutors to this homework question now:have to: (a) Write the claim mathematically and identify H0 and Ha (b) Find the standardized test statistic Z and its corresponding area. (c) Find the P-value. (d) Decide whether to reject or fail to reject the null hypothesis. (e) Interpret the decision in the context of the original claim.
2 answers