Question

62.3 62.3 correct 67.8 67.8 incorrect 61.8 61.8 incorrect 68.4 68.4 correct At least one of the answers above is NOT correct. (1 point) In a simple random sample of 1120 full-time students from a large population, it is found that 729 study, on average, more than 80 minutes per day. Use this sample to estimate the true percentage, p, of full-time students in this population who study, on average, more than 80 minutes per day. In each question below, give answers as percents, rounded to at least two decimal places. (a) Find a 90% confidence interval for p: 63.3 % <p < 67.4 (b) Find a 95% confidence interval for p: 62.3 % <p < 67.8 olo (c) Find a 98% confidence interval for p: 61.8 % <p < 68.4

Answer 1

(a)

Sample size = n = 1120 Sample number of events = x = 729 Sample proportion = p = x/n = (729/1120) = 0.6509 (point estimate of population proportion) Stanadard Error = P(1-0) ? = sqrt(0.6509*(1-0.6509)/1120) = 0.0142 V n Confidence Level = 90 Significance Level = a = (100-90)% = 0.1 Critical value = z* = 1.645 [Using Excel =ABS(NORMINV(0.1/2,0,1))] Margin of Error = z*(standard error )= 1.645*0.0142 = 0.0234 Lower Limit =- Margin of Error = 0.6509 -0.0234 = 0.6275 Upper Limit = + Margin of Error = 0.6509 +0.0234 = 0.6743

Ans : 62.75 % <p < 67.43 %

(b)

Sample size = n = 1120 Sample number of events = x = 729 Sample proportion = p = x/n = (729/1120) = 0.6509 (point estimate of population proportion) Stanadard Error = P(1-) - = sqrt(0.6509*(1-0.6509)/1120) = 0.0142 ✓n Confidence Level = 95 Significance Level = Q = (100-95)% = 0.05 Critical value = z* = 1.96 [Using Excel =ABS(NORMINV(0.05/2,0,1))] Margin of Error = z* (standard error )= 1.96*0.0142 = 0.0279 Lower Limit =- Margin of Error = 0.6509 -0.0279 = 0.623 Upper Limit =p + Margin of Error = 0.6509 +0.0279 = 0.6788

Ans : 62.30% <p < 67.88 %

(c)

Sample size = n = 1120 Sample number of events = x = 729 Sample proportion = p = x/n = (729/1120) = 0.6509 (point estimate of population proportion) Stanadard Error = P(1-0) ? = sqrt(0.6509*(1-0.6509)/1120) = 0.0142 ✓n Confidence Level = 98 Significance Level = a = (100-98)% = 0.02 Critical value = z* = 2.326 [Using Excel =ABS(NORMINV(0.02/2,0,1))] Margin of Error = z* (standard error )= 2.326*0.0142 = 0.0331 Lower Limit =P - Margin of Error = 0.6509 -0.0331 = 0.6178 Upper Limit =p + Margin of Error = 0.6509 + 0.0331 = 0.684

Ans : 61.78 % < p < 68.40 %