Question

In the planning stage, a sample proportion is estimated as pˆ = 72/80 = 0.90. Use this information to compute the minimum sample size n required to estimate p with 95% confidence if the desired margin of error E = 0.05. What happens to n if you decide to estimate p with 90% confidence? (You may find it useful to reference the z table. Round intermediate calculations to at least 4 decimal places and "z" value to 3 decimal places. Round up your answers to the nearest whole number.)

	Confidence Level n 95% 90%

Answer 1

p) let los be d.o.o5 61196 (from e- table) Zd/2 70.025 and given E 0.05 E 0.90 (i* pci-p) (1.26.) 8-92.00) n = no 138.2994 m 138 be a: 0-10 (1) let sal - 1.645 (from Z - table) Zar Zo.o5 mo 63605) 70-9)01-0.4) mo 97.4169 mŅ 1977 as the confidence level decreases the required sample size decreases.