A parametric test requires the following conditions
a null hypothesis that can be stated in terms of a parameter
a meaningful level of measurement has been achieved that gives validity to difference
a test statistic that follows a known distribution
all of the above is true
A non-parametric test requires
a null hypothesis
the calculation of a test statistic
the use of critical values
all of the above
A chi-squared test statistic of 16.548 has been calculated for a 4 by 5 cross tabulation (with all expected frequencies above 5). Using a 5% significance level
you would compare with a critical value of 21.0 and not reject the null hypothesis
you would compare with a critical value of 21.0 and reject the null hypothesis
you would compare with a critical value of 14.8 and not reject the null hypothesis
you would compare with a critical value of 14.8 and reject the null hypothesis
A frequency table has been used to report the length of life of a component. The table has 8 intervals and has been used to calculate the sample mean and sample standard deviation. It has been decided to test at the 1% significance level whether the length of life follows a normal distribution. A chi-squared test statistic of 13.246 was calculated by comparing the observed and expected frequencies (with all expected frequencies above 5). You would compare the chi-squared value with a critical value of
11.1 and not reject the null hypothesis
11.1 and reject the null hypothesis
15.1 and not reject the null hypothesis
15.1 and reject the null hypothesis
You are using the Mann-Whitney U test for a small sample and have been given the following summary:
you would compare a T value of 10 with a critical value of U of 6 and reject null hypothesis
you would compare a T value of 10 with a critical value of U of 6 and not reject null hypothesis
you would compare a T value of 10 with a critical value of U of 6 and reject null hypothesis
you would compare a T value of 10.5 with a critical value of U of 12 and not reject null hypothesis
You are using the Mann-Whitney U test for a large sample and have been given the following statistics: T = 258, n1 = 15 and n2 = 25. Using a 5% significance level you would compare a z value of
+1.87 with +1.645 and reject null hypothesis
+1.87 with +1.96 and not reject null hypothesis
-2.87with +1.96 and reject null hypothesis
+2.87 with +1.96 and not reject null hypothesis
A company has identified a number of fraudulent claims. To test whether they are random they should use a: