Question

According to the Canadian Internet Use Survey conducted by Statistics Canada in 2018, 74% of Canadian seniors (65 and over) used the Internet on a daily or near-daily basis. A market researcher wishes to estimate, the current (Summer 2020) proportion of all Canadian seniors who use the Internet on a regular basis. (a) How large should the sample size be so that a 96 % confidence interval for the proportion of all Canadian seniors who use the Internet on a regular basis has a margin of error of 0.02? Consider the result of p = 0.74 from the 2018 study as the preliminary sample proportion. (Note: Enter the whole number value without any symbol or equation) (b) It is expected that Internet use among Canadian seniors has grown steadily since 2018. So, the proportion of users obtained from the 2018 survey, will not be valid in 2020. Assuming the current/valid sample proportion is unknown, what is the most conservative estimate of sample size that would limit the margin of error to within 0.02 of the population proportion for a 96 % confidence interval? (Note: Enter whole number value without any symbol or equation.) © Should the sample size obtained in part (6) bigger than the sample size obtained in part (a)? (Note: Enter 1 if Yes, 2 if No).

Answer 1

a)

here margin of error E =	0.020
for96% CI crtiical Z          =	2.054	from excel:normsinv((1+0.96)/2)
estimated prop.=p=	0.740
sample size n= p*(1-p)*(z/E)2=	2029	(rounding up)

b)

here margin of error E =	0.020
for96% CI crtiical Z          =	2.054	from excel:normsinv((1+0.96)/2)
estimated prop.=p=	0.500
sample size n= p*(1-p)*(z/E)2=	2637	(rounding up)

c)

1