At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Since the test statistic is not greater than the critical value, the null hypothesis cannot be rejected. Therefore, the claim that the mean score is 71 is not rejected.
1. We are given the population of statistics final exams and asked to test the claim that the mean score is 71 using an alternative hypothesis that the mean score is different from 71. We are also given the sample statistics including the sample size (n = 26), the sample mean (ã = 72), and the sample standard deviation (s = 18). We are asked to use a significance level of α = 0.05.
2. Since we are assuming that the population is normally distributed, we will use a two-tailed z-test to test the claim.
3. The null hypothesis is that the mean score is 71 and the alternative hypothesis is that the mean score is different from 71.
4. The test statistic is calculated as:
Test statistic = (sample mean - population mean) / (standard deviation/√n)
= (72 - 71) / (18/√26)
= 0.283
5. The critical values for a two-tailed z-test with a significance level of α = 0.05 are 1.645 for both the positive and negative critical values.
6. Since the test statistic (0.283) is not greater than the critical value
Learn more about null hypothesis here
https://brainly.com/question/28920252
#SPJ4
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
