In a Paired t test conducted for a sample of 11 data items, what should the Degree of freedom be?
OPTIONS 20 18 11 10 ANSWER 10 EXPLANATION Degree of freedom for a paired t test is n – 1, so 10 is correct.
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Tag: LSSBB Exam
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If p-value of a statistical 2 Sample t test conducted by a Black Belt is 0.032 at a Significance level of 0.05, what is the Degree of confidence with the Black Belt with which he can reject the conclusion that there is no statistically significant difference between the two groups?
OPTIONS 0.95 0.9 0.968 0.96 ANSWER 0.968 EXPLANATION P-value of 0.032, 3.2% of risk, which means with 96.8% confidence, the Black Belt can reject the null, so 0.968 is correct.
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If p-value is 0.103 at an α of 0.05 for a 2 Sample test for a two-tailed hypothesis, what should the Black Belt deduce
OPTIONS The statistic of both the groups don’t differ significantly The statistic of both the groups differ significantly The statistic of both the groups are found of a mixed level None of the above ANSWER The statistic of both the groups don’t differ significantly EXPLANATION...
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If a Black Belt wishes to study a metric that shows the statistical accuracy of a test, which of these metrics should he choose
OPTIONS Type III error Type II error Power Type I error ANSWER Power EXPLANATION Power indicates accuracy of a test, so POWER is correct.
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If Confidence Level is 90%, what should the Black Belt assume about his statistical test?
OPTIONS The test will have a high chance of errors The risk in the experiment is low The probability of committing a Type I error is high The risk in the experiment is high ANSWER The probability of committing a Type I error is high...
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At a significance level 5%, a Black Belt committed rejected the fact that the data belongs to a normal distribution, when in reality it belonged to the distribution. What did the Black Belt do?
OPTIONS He committed a Type I error He committed a Type II error He committed a Type III error None of the above ANSWER He committed a Type I error EXPLANATION The condition described in the question best meets the description of a Type I...
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For a data meeting normality conditions, which of the tests mentioned below would you use to test if the data comes from the population assuming sample size is 25?
OPTIONS Z testF testT testANOVA ANSWER T test EXPLANATION Typically for sample size less than 30, one would use the T test.
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For a class of 60 reporting an approximate Mathematics score of 50%, the confidence intervals for the Mathematics score at 95% level is
OPTIONS 36%, 64% 32%, 68% 38%, 62% 40%, 60% ANSWER 38%, 62% EXPLANATION Using standard confidence intervals formula, 38%, 62% is correct.
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For determining confidence intervals for Variances, which one of the below distributions would you popularly use?
OPTIONS Normal Distribution T distribution Chi-Square Distribution Binomial ANSWER Chi-Square Distribution EXPLANATION Chi-Square distribution is typically used for determining confidence intervals for variances, so Chi-Square Distribution is correct.
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If the confidence intervals for a given experiment is (10,12), what is the probability that 8.5 will fall in the range?
OPTIONS 0.025 0.0025 0.02 0.05 ANSWER 0.025 EXPLANATION Using Chi-Square confidence intervals, the correct answer is 0.025.
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