please make response clear and no excel or other program please. 23-25 Given the
ID: 3321697 • Letter: P
Question
please make response clear and no excel or other program please.
23-25 Given the following information about a hypothesis test of the difference between two means based on independent random samples: samples are obtained from normally distributed populations having equal variances 23. What is the cquation for the test statistic? Assume that the 24. What is the calculated value of the test statistic? A. t 2.48 B. t 1.5 C. t 2.823 D. t 1.674 E. t 1.063 25. What is the critical value of the test statistic at a significance level of 0.05? 26. The chi-square goodness of fit test can be used when: A. We conduct a binomial experiment B. We conduct a multinomial experiment C. We perform a hypothesis test to determine if a population has a normal distribution. D. We perform a hypothesis test to determine if two population variances significantly differ from each otherExplanation / Answer
Question 23
Test statistic formula for pooled variance t test is given as below:
t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]
Where Sp2 is pooled variance
Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)
Question 24
We are given
X1bar = 12
X2bar = 9
S1 = 5
S2 = 3
n1 = 13
n2 = 10
df = n1 + n2 – 2 = 21
= 0.05
Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)
Sp2 = [(13 – 1)*5^2 + (10 – 1)*3^2]/(13 + 10 – 2)
Sp2 = 18.1429
(X1bar – X2bar) = 12 – 9 = 3
t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]
t = 3 / sqrt[18.1429*((1/13)+(1/10))]
t = 3/1.7916
t = 1.6745
Correct Answer = D. t = 1.674
Question 25
df = n1 + n2 – 2 = 21
= 0.05
Critical value = -2.0796 and 2.0796
Question 26
Answer: C. We perform a hypothesis test to determine if a population has a normal distribution.
(Because, we use chi square goodness of fit test for testing whether the data follows a specific distribution or not.)
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