AS Description:This is the 4th and final portion of the project. You will continue to work with the same variable in the previous two portions. However, now you will now go back to your data to analyze this variable even more by conducting a one way ANOVA.
AS Instructions:Complete and answer the following:
1. Determine how many populations you will be comparing and set up the null and alternative hypothesis
2. Find the test statistic (F stat) and the P value
3. Use alpha = .05 to conclude whether the means are the same or different
Choice of variable Choice of variable Here, we will chose our variable of interest. The choice of the variable is from the regression project data That was used earlier. Description of variable We have chosen IQ score variable as our variable of interest. The reason behind the choice of the variable is that we want to discover the descriptive statistics of this variable. As compare to other variable “Video game hours” this variable is an important variable. Also, it is the dependent variable in the regression study. Sample Mean Calculation of mean: Sample mean = Sum of all values / total observations Sample mean = 51.32 Calculation of sample standard deviation Sample standard deviation = 17.4014 95% Confidence interval Sample mean (x-bar)= 51.32 Sample standard deviation (s) = 17.401 Critical value at 95% = 1.96 Sample size, n = 25 Thus, Confidence interval = (x-bar ± 1.96 *s / ) Confidence interval = ( 51.32 ± 1.96*17.40 / ) Confidence interval = (44.50, 58.14) Interpretation of confidence interval The 95% confidence interval is (44.50 , 58.14). Thus, it can be concluded that we are 95% confident that the sample mean of IQ score lies between 44.50 and 58.14. Hypothesis Test about Mean Hypothesis Test about Mean Aim: Here we apply t-test to test our hypothesis Choice of Variable We will choose our variable of interest as dependent variable. The variable is the IQ score. The data set is given in the table. IQ Score61568175545373536158647473 663421314438284826314040 Hypothesis question Here, we want to test whether the average IQ score of the students is different from 50. Here, 50 is our population mean. The level of significance for test is 0.05. The hypotheses are given as follows: The test is a two-tailed test. Descriptive Statistics The sample mean of the IQ score, x-bar = 51.32. The sample standard deviation, S = 17.40. The sample size, n = 25 Test-statistics The test-statistics is given as follows: Test-statistic, t = The test-statistic is 0.38 Conclusion and p-value The p-value is given as follows: P-value = 0.71 Conclusion: The p-value for the test is 0.71. Which is greater than the level of significance 0.05. Thus, the null hypothesis is not rejected. Hence, we concluded that the IQ score is equal to 50. Since, the null hypothesis is not rejected. So, there are chances that we may committed type-2 error. That is if original IQ score is really different from 50. End THANK YOU 0 1 :50 :50 H H m m = ¹ / x sn m - 51.3250 17.40/25 0.38 - = =