Use the following information for questions 17 to 18 In a clinical trial, 50 pat

Question

Use the following information for questions 17 to 18 In a clinical trial, 50 patients who received a new medication are randomly selected. It was found that 10 of them suffered serious side effects from this new medication. let p denote the population proportion of patients suffered serious side effects from this new medication. 17. Find a point estimate for p and also construct a 95% confidence interval for p. (a) 20, (19.889, 20.111) (b) 0.80, (0.689, 0911) (c) 0.20, (0.089, 0.311) (d) 20, (0.145, 0.255) (e) 0.20, (0.107, 0293) Using the proportion estimate from the above sample, find the minimum sample size needed to estimate the population proportion P with 90% confidence. The estimate must be accurate to within .04 of p 18. (a) 100 (b) 164 (c) 385 (d) 271 (e) 423 For questions 19 and 20: Measurement of the waiting time of 24 randomly selected patients at a hospital emergency room gave mean and sample standard deviation 11.3 and 6.5 minutes. Assume that the waiting times are normally distributed 19, A 95% confidence interval for the mean waiting time of all patients is about (a) 11.3± (1.645)( (b) 11.3± (1.96) ) ) (c) 11.3± (1.714)( ) (d) 11.3± (2.069)( (e) 11.3±(2.575)(%) 20. If, after you obtain the confidence interval, you find it to be too wide, which of the following remedial steps can you take to reduce the width of the confidence interval? I. To construct a 90% confidence interval instead of a 95% one. 11, To construct a 99% confidence interval instead of a 95% one. 111 To re-do the 95% confidence interval with only a half of the sample data. (a) I only (b) only (c)Ionl (d) and I only (e II and 1 only

Explanation / Answer

Question 17

point estimate p = 10/50 = 0.20

Confidencenterval 95% = p^ +- Z95% sqrt [p * (1-p) /N]

= 0.20 +- 1.96 * sqrt [0.20 * 0.80/50]

= 0.20 +- 0.11087

= (0.089, 0.311)

Question 18

Let say the required sample size = n

Here margin of error = 0.04 = critical test statistic * standard error of proportion

standard error of proportion = sqrt [ 0.20 * 0.80/n] = 0.4/sqrt(n)

So,

0.04 = 1.96 * 0.4/ sqrt(n)

sqrt(n) = 1.96 * 0.4/0.04 = 19.6

n = 384.16 or 385 option C is correct.

Question 19

95% confidence interval = x +- t23,0.05 (s/ sqrt(n)

here dF = 23 and alpha = 0.05 ; t23,0.05 = 2.069

so option C is correct here.

Quetion 20

(i) Decrease confidence level to 90% will reduce the width of confidence interval.

(ii) Reducing sample size to half will icrease the width of confidence interval.

So only option A is correct

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