Question

The Ohio Department of Education maintains records of average number of years of teaching experience for each public school in the state. During the 2012-2013 school year, it was reported that the average number of years of teaching experience at Ohio high schools was 14.3 years. Suppose that an intern working in educational policy research wants to determine whether the average number of years of teaching experience of teachers in Ohio high schools changed between the 2012-2013 and 2013-2014 school years The intern selected a random sample of 14 high schools, and average number of years of teaching experience for the 2013- 2014 school year at each of these 14 schools is recorded below. Prior years' data suggest that mean teaching experience at Ohio public high schools is normally distributed. 19,8, 12,7, 13, 14, 15, 11, 16, 14, 17,9, 12,12 If you wish, you may download the data in your preferred format CrunchIt! CSV Excel JMP Mac Text Minitab PC Text R SPSS TI Calc Use a two-tailed one-sample 1-test to determine whether average number of years of teaching experience at Ohio high schools during the 2013-2014 school year was different from 14.3 years. Have the requirements for a one-sample t-test been met? If they have not been met, leave the remaining questions blank. No, the intern selected a random sample from a normally distributed population, but his sample is to0 small No, the intern selected a normally distributed random sample, but he does not know the population standard deviation Yes, the intern selected a random sample from a normally distributed population, and his sample contains no outliers Yes, the intern selected a random sample that is normally distributed and contains no outliers

Answer 1

19
8
12
7
13
14
15
11
16
14
17
9
12
12
12.78571	average
3.400873	sd

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 14.3 Ha: u # 14.3 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two- tailed test is te = 2.16. The rejection region for this two-tailed test is R = {t:t > 2.16} (3) Test Statistics The t-statistic is computed as follows: X - Mo s/m 12.7857 – 14.3 3.40087/14 = -1.666 (4) Decision about the null hypothesis Since it is observed that t = 1.666<te=2.16, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.1196, and since p=0.1196 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 14.3, at the 0.05 significance level.

requirements

option C) is correct

Hypothesis
option D)
t = -1.666
p-value = 0.1196
option D) 0.10 < P < 0.20

The intern should fail to reject his null hypothesis , there is not   evidence