Using the data below and a 0.05 significant level test the claim that the responses occurred with percentages of 15% 20% 25% 25% 15% respectively determine the test statistic and critical value
To test the claim, we can use a chi-squared goodness-of-fit test. The null hypothesis is that the observed responses occur with frequencies corresponding to the claimed percentages.
Observed frequencies:
15%: 3
20%: 1
25%: 2
Expected frequencies based on claimed percentages:
15%: 15% of 6 = 0.15*6 = 0.9
20%: 20% of 6 = 0.2*6 = 1.2
25%: 25% of 6 = 0.25*6 = 1.5
Now, let's calculate the chi-squared test statistic:
Chi-squared = Σ((Observed - Expected)^2 / Expected)
Chi-squared = ((3-0.9)^2 / 0.9) + ((1-1.2)^2 / 1.2) + ((2-1.5)^2 / 1.5)
Chi-squared = (2.1^2 / 0.9) + (0.2^2 / 1.2) + (0.5^2 / 1.5)
Chi-squared ≈ 4.67
Now, we need to determine the critical value of the chi-squared distribution with 2 degrees of freedom (k-1 categories - 1), which corresponds to a significance level of 0.05. Consulting a chi-squared distribution table, the critical value is approximately 5.99 for 2 degrees of freedom.
Since the test statistic (4.67) is less than the critical value (5.99), we fail to reject the null hypothesis. This means that there is not enough evidence to reject the claim that the responses occurred with percentages of 15%, 20%, 25%, 25%, and 15%.