Question

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Q1 A school found that 78 out of a random sample of 100 students had read a book in a month. (a) Find a 96% confidence interval for the proportion in the corresponding population. (b) The school has decided that a more precise estimate is needed and wishes to reduce the width of the confidence interval by a quarter. How many more students should be investigated if the school wants to find the 96% confidence interval for true proportion later on?

Answer 1

TOPIC:Confidence interval for the population proportion and the required sample size.

the alx a) Define n = sample p = sample book -ñ size = 100. students proporation of reading in a month." - -0.78. Leb, p = Corresponding population proportion. Heres & 2 0.04, dos 96% confidence level. confidence level = 96%.] => 100 (1-2% - 96 2) 220.04. Now, E = margin of error. upper / ron] =) 2 (zal x Zal2 point of NOOD 0.78 xo.22. 100 1.02 20.02 = 2.0537 0.78 0.22 X 2.0537 100 From z value -critical table, = 0.0851.

Thus, the required 96 7. CI for the population proportion (p) is = (6-E, P+E). - (0.78 -0.0661, 0.78 + 0.085). (0.6949, 0.963), ~ (0.695, 0.865 in three decimals (Anss by Ret, m= Required sample size to reduce the width of the confidence interval by a quarter. The previous width of the confidence interval - (0.865-0. 695). 20. 170. = Wo, say The new width of the confidence interval (WN) = wo - 1 / 4 Wo. 3 . Reduced by sy Wo. La quarter. - 0.110. = 0.1275

Thus, new margin of erron (EN) = WN 0.1275 Here & Preliminary bepulation But, we have = 0.06375. estimate of the proportion = 0.78 => End bombe X Zay, 20-66375, Zal2 = 20102 10. 78 X 0.22 x 2.0537 = 0.06375. = 2:0537 found before >) ma - 0.78 X0. 22 X20537) 2 (0.06 375) 2 = m = 178. 086 a) m w/ 178 ( Rounded to the) nearest integer They need to sample (178-100) = 178 more students to reduce the width of the confidence interval by a quarter. (Ans).