In: Statistics and Probability
(Questions 1 and 2) An experiment is conducted. The sample size is 30. A sample average is observed. To test the null hypothesis that the population average equals a particular value versus the alternative hypothesis that the population average is not equal to a particular value, a two tailed hypothesis test is performed. (Critical region in the two tails). The test statistic— the sample average—is normally distributed. The stated significance levels for the test are 5% and 1%. In other words, the null hypothesis is rejected at a stated significance level of 5% or 1%.
2. (Use the z table to answer this question) If the observed value of the z statistic is -2.10, the observed significance level, (the p value) is:
(Question 3) A researcher wishes to test the null hypothesis that a die is fair against the alternative hypotheses that a die is not fair. A die is rolled 360 times. There are 6 categories. The observed value of the chi square statistic is 10.23.
3. (Use the chi square table to answer this question) In her written report of the results of the experiment, the researcher should declare,
1) Answer: Reject the null hypothesis at the 5% level of significance but not the 1% level of significance
Explanation:
The critical value at 5% level of significance = -1.96
The critical value at 1% level of significance = -2.58
The observed value of z statistic is less than-1.96 but greater than -2.58. Hence we reject the null hypothesis at the 5% level of significance but not the 1% level of significance
2) Answer: None of these
Explanation:
P-value = 2*P(Z < -2.10)
=2*(0.0179)
P-value =0.0358
P-value = 3.58%
3) Answer: The result is not significant at the either the 1% level or the 5% level.
Explanation:
The degrees of freedom for Chi-square test = k-1 = 6-1=5
The critical value of chi-square at 5 df and 5% level of significance = 11.07
The critical value of chi-square at 5 df and 1% level of significance = 15.09
The value of chi square statistic is less than both critical values. Hence we conclude that the result is not significant at the either the 1% level or the 5% level.