Question

Question 32 3 pts Proportion Problem. Each year Delta Dental does a survey of parents concerning the tradition of the Tooth Fairy. A random sample of 1,058 parents were asked about how much money they leave for a tooth and questions concerning how their child responded to the concept of the Tooth Fairy. In one question, 508 parents indicated that their child saved the money they got for their tooth. You are asked to Test to see if this proportion is different from .5. Use an alpha level of.05. What is the Critical Value used for this test? I just want the answer. Use 4 decimal places for your answer and use the proper rules of rounding. - 1.96 Question 33 3 pts Proportion Problem. Each year Delta Dental does a survey of parents concerning the tradition of the Tooth Fairy. A random sample of 1,058 parents were asked about how much money they leave for a tooth and questions concerning how their child responded to the concept of the Tooth Fairy. In one question, 508 parents indicated that their child saved the money they got for their tooth. You are asked to Test to see if this proportion is different from.5. Use an alpha level of .05. What is the Standard Error for this test? I just want the answer. Use 4 decimal places for your answer and use the proper rules of rounding.

Answer 1

Here we have х x 508, n 1058, p -0.4802 n Hypotheses are: H p 0.5 Ha p 0.5 Level of significance: a 0.05 Test is two tailed The standard deviation is: P(1-р) -sqrt((0.5*(1-0.5))/1058) п 0.0154 Test Statistics: (0.4802-0.5)/(0.0154) =-1.29 Critical value: CV = 1.96 Rejection Region: If z > 1.96, Reject HO Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis P-value P-value-0.1971 Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis Excel function for critical value: =NORMSINV(1-(0.05/2)) -2*(1-NORMSDIST(1.29)) Excel function for p-value:

Answers:

32: CV = 1.9600

33: SE = 0.0154

------------------------------------------------------------

28:

Here we have 1491.43, s =411.53, n =14 Hypotheses are: Ho:u= 1327 Н: и> 1327 aa 0.05 Level of significance: Test is one tailed (right tailed) df n-1-13 Degree of freedom: Test Statistics t = = (1491.43-1327)/(411.53/sqrt(14)) s/Nn = 1.495

---------------------------------------------------------------

Answer 7: False

Answer 8: True

------------------------------------------------------

29:

Here we have T1491.43, S =411.53, n =14 Hypotheses are: = 1327 0 Н: и> 1327 Level of significance: aa 0.05 Test is one tailed (right tailed) df n-1-13 Degree of freedom Test Statistics: (1491.43-1327)/(411.53/sqrt(14)) s/Vn = 1.495 Critical value: 1.771 If It >1.771, Reject HO Rejection Region: Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value 0.079 P-value: Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: =TINV(0.1,13) =TDIST (1.495,13,1) Excel function for p-value: