A group of scientists wanted to compare the intelligence of the Sus Scrofa Domes
ID: 3227197 • Letter: A
Question
A group of scientists wanted to compare the intelligence of the Sus Scrofa Domesticus (SSD) and the Caris Lupus (CIF) After a series of tests. they gathered the average scores of a sample of 15 SSD and 16 CLF. The SSD average intelligence score was 60 points (variance of 36 points) and an average intelligence score of 56 points (variance of 25 points) for the CLF. The scientists want to know with a ninety percent certainty if these two animals different intelligence levels a Write the Null and Alternative Hypothesis b. What is alpha (a) in this scenario c. Would you conduct 1-tailed test (upper or lower) or a 2-tailed test? d. Calculate the degrees of freedom for this test e. What critical value will you use to identity rejection regions? f. Calculate the pooled variance for these samples g. Write the test statistic formula you should use h. Calculate the test statistic i. State your conclusion - Reject or fail reject the null Hypothesis j. What should the scientists report per the results of this test? The scientists also want to know if there is a difference in the variances. Using the information problem 4 answer the following: a. Write the Null and Hypothesis b According to H_a and the a used in problem 4, which distribution table should you use? c. Which sample has the higher variance? d. What are the degrees of freedom for the sample with the higher variance? e. What are the freedom for the sample the lower variance? f. Which one is your numerator degrees of freedom? g. Which one is your denominator degrees of freedom? h. What is the value to identify the rejection region? i. Calculate the test statistic j. Do you reject or fail to reject the null hypothesis? k. What can scientists infer based on the test result. To corroborate the results from the test done in problem 4. the scientists collected a larger sample and the new descriptive statistics are reported in the table below. Using the same a and hypotheses in problem 4, answer the following a. What critical value will you use to identify the rejection regions? b. Write the test statistic formula you should use c. Calculate the test statistic d. State your conclusion-Reject or fail to reject the null hypothesis e. What should the scientists report per the results of this test? f. What is the p-value? g To what should you compare the p-value to make a conclusion on this hypothesis test? h. What's your conclusion based on the p value and how did you decide? Based on the information in problem 5, the scientists want to know the ninety-five percent confidence interval of the population difference in means. a. Write the appropriate confidence interval formula b. Calculate the confidence interval The scientists want to see if, within 3.2 points, their samples in problem 5 were large enough for their objectives. Previous data states that the population standard deviations of the groups are l4 points and 13 points. a. Write the required sample size formula b. Calculate the required sample size c. Did the scientists use a large enough sample in problem 5?Explanation / Answer
4)
Null Hypothesis H0: The intelligence score of SSD is equal to the intelligence score of CLF
Alternative Hypothesis H1: The intelligence score of SSD is not equal to the intelligence score of CLF.
alpha = 1 – 0.9 = 0.1
As we are testing for unequal condition in the alternative hypothesis, it is a two-tail test.
Degree of Freedom, DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
= (36/15 + 25/16)2 / { [ (36 / 15)2 / (15 - 1) ] + [ (25 / 16)2 / (16 - 1) ] }
= 27 (rounded to nearest integer)
Critical value of rejection region – The value of t at df = 27 and significance level = 0.1/2 = 0.05 is t < - 1.703 or t > 1.703
Pooled variance = ((n1-1)s1^2 + (n2-1)s2^2)/(n1+n2-2)
= ((15-1)36 + (16-1)25)/(16+15-2)
= 30.31
Pooled standard error = sqrt(pooled variance * (1/n1 + 1/n2)))
= sqrt(30.31 * (1/16 + 1/15)) = 1.979
Test statistic formula, t = (Difference in means)/Pooled standard error
Test statistic, t = (60-56)/1.979 = 2.021
Conclusion – As t value is greater than critical value (1.703), we reject the null hypothesis and conclude that that the intelligence score of SSD is not equal to the intelligence score of CLF.
Scientist report – There is statistical evidence that the intelligence level of SSD is different than the intelligence score of CLF.
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