At a 90% confidence level, we conclude that your class is better than average (83). The expert determines the average grade for all people who take a class is 83.
The manager of the bank decides to do a survey among the customers of his bank and takes a simple random sample of 200 customers aged 18 and over.
What is the value of the test statistic? (Round your answer to 2 decimal places.) Value of the test statistic=(21-20)/[4/sqrt(49)]=1.75 d. What is your decision regarding H0?
Significance Level, α=0.10 Using sample data, we will conduct a paired sample t-test to test the null hypothesis.
Compare the test statistic to the rejection region and make a judgment about the null and alternative hypotheses. 5. Interpret the statistical decision in terms of the problem. 6.
What we do all the time is to prove or disprove the null. d) Provide an example of a hypothesis test you could conduct at work. What is the measure that you will test? Ho: men and women earn the same income and Ha: men earn more than women.
This solution consists of details of using one sample t-test to test the claim that the average activating temperature is at least 135 degrees F.
Also, the distribution of the data should be considered. Step 4: Select an error level. A researcher must determine the p-level for the test.
The significance level is 0.10. From the above result, since P-value is 0.1655 which is large than 0.10, we fail to reject null hypothesis. Hence, the conclusion is that there is no difference between new fertilizer and old fertilizer.
H0:p1-p2=0 Null hypothesis Ha:p1-p2≠0 Alternative hypothesis α₌0.10 Since the test is to be carried out at the 0.10 level of significance, the critical values are -1.645 and +1.645.