is Question: 1 pt 13 of 40 (11 complete) This Test: 40 pts possibk The walting t
ID: 3074964 • Letter: I
Question
is Question: 1 pt 13 of 40 (11 complete) This Test: 40 pts possibk The walting tmes (in minutes) of a random sample of 22 people at a bank have a sample standard deviation of 3.3 minutos Construct a confidence interval for the optat on vara ce the populair standard de ai n Use a 90% level of con den e Ass me the sample is What is the contfidence interval for the population variance o Round to one decimal place as needed.) Interpret the results. Select the correct choice below and 1 in the answer boxles) to complee your choice anomaly dst but d Round to one decimal place as needed) OA with 90% cord de ce, you can say that the poplation var ance s ben een Oawn scotance, you can say that he population variance is guter than weh 10% oc with 10% ort one you can say th at the population var ance is less than Oo oferce you can say hat the podatin mresbeta en What is the confidence interval for the population standand deviation ? (Round to one decimal place as needed) Interpnet the nesults. Solect the correct choice below and fil in the ansaver bos(es) to complete your choice Round to one decimail place as needed) OA. with 10% o f de ce you can saythat the pop it sa dard deviatin it OS wtht0% coteren you can say fat the pop i to stand ddivus no c, wth 9% o to ce.yucan say that the pop daton sindeddeutn s ith90% c teen you an say that ro oli stanu d de knis and Click to select your answerts). 6 8 0 TPoupExplanation / Answer
Solution :- Given that n = 22, standard deviation = 3.3 90% confidence interval
df = n-1 = 21
X^2R = 32.67 , X^2L = 11.59
The confidence interval for the population variance ^2 =
=> (n-1)*s^2/X^2R < ^2 < (n-1)*s^2/X^2L
= (21*3.3^2)/32.67 < ^2 < (21*3.3^2)/11.59
= 7 < ^2 < 19.7
=> option A.
=> The confidence interval for the population standard deviation =
=> sqrt((n-1)*s^2/X^2R) < < sqrt((n-1)*s^2/X^2L)
= sqrt(7) < < sqrt(19.7)
= 2.6 < < 4.4
=> option D.
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