3. (from Q30, P. 529-530) Testing the ads. The company in the above exercise con

Question

3. (from Q30, P. 529-530) Testing the ads. The company in the above exercise contacts 600 people selected at random, and only 133 remember the ad. a. (2 marks) Calculate a 95% confidence interval for the proportion of all residents who remember the ad. b. (2 marks) Calculate the P-value for testing the hypotheses given in the last question. c. (2 marks) Should the company renew the contract? Use the confidence interval and P-value to justify your conclusion. 4. (from Q6, P. 580, Intro Stats) Surgery and Germs Joseph Lister (5 April 1827- 10 February 1912 and for whom Listerine is named!) was a British physician who was interested in th role of bacteria in human infections. He suspected that germs were involved in transmittin infection, and so he tried using carbolic acid as an operating room disinfectant. In 75 amp tations, he used carbolic acid 40 times. Of the 40 amputations using carbolic acid, 34 of ti patients lived. In the 35 amputations without carbolic acid, 19 lived. The question of intere is whether carbolic acid is effective in increasing the chances of surviving an amputation. a. (1 mark) Is this an experiment or an observational study? b. (4 marks) Es timate the difference in survival rates using a 95% confidence interval interpret your interval in context. c. (2 marks) What reservations do you have about the design of this study?

Explanation / Answer

Answer to question# 4)

Group I:

Carbolic acid

n1 = 40

x1 = 34 survive

Group II:

Non carbolic acid

n2 = 35

x2 = 19 suvived

.

p1 = 34 /40 = 0.85

P2 = 19/35 = 0.54

SE = sqrt(0.85*0.15/40 + 0.54*0.46/35)

SE = 0.1014

.

part a)

There has been an intervention of carbolic acid as room freshner. Thus this is an experimental study and not an observational study. In observational study there is niether any intervention nor any control on the variables in the study. The main aim of this study is to observe the effect of the intervention of corbolic acid on the survival rate

.

part b)

Formula of confidence interval is:

(p1^-p2^) - z* SE , (p1^-p2^) + z*SE

.

We got:

p1^ = 0.85

P2^ = 0.54

SE = 0.1014

The value of Z for 95% confidence level is 1.96

.

On plugging these values we get:

(0.85 -0.54) - 1.96 * 0.1014 , (0.85 - 0.54) +1.96 * 0.1014

0.1113 , 0.5087

.

Interpretation; Since the comfidence interval of the difference of proportions do not contain the value 0 , this implies that the difference is proportions is significant, and since both the value sof proprotions is positive this implies proportion of survival with carbolic acid (p1^) is higher than the proportion of survival without carbolic acid (p2^)

.

Part c)

The only reservation in this design is that the cause of the increased survival rate may not be solely due to the presence of carbolic acid. There may be many extraneous factors which have not been considered here. There has been no control on the other important factors on which the survival rate may depend.

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