Question

Last summer, Survey USA published results of a survey stating that 315 of 563 randomly sampled Kansas residents planned to set off fireworks on July 4th. Round all results to 4 decimal places.

1. Calculate the point estimate for the proportion of Kansas residents that planned to set off fireworks on July 4th

2. Calculate the standard error for the point estimate you calculated in part 1.

3. Calculate the margin of error for a 90 % confidence interval for the proportion of Kansas residents that planned to set off fireworks on July 4th.

4. What are the lower and upper limits for the 90 % confidence interval.

( ,  )

5. Use the information from Survey USA poll to determine the sample size needed to construct a 95% confidence interval with a margin of error of no more than 3.1%. For consistency, use the reported sample proportion for the planning value of p* (rounded to 4 decimal places) and round your Z-value to 3 decimal places. Your answer should be an integer.

Answer 1

Solution :

Given that n = 563, x = 315

1.
=> point estimate of the proportion p = x/n

= 315/563

= 0.5595

2.
=> standard error Se = sqrt(p*(1-p)/n)

= sqrt(0.5595*(1 - 0.5595)/563)

= 0.0209

3.
=> for 90% confidence interval, Z = 1.645

=> margin of error E = Z*Se

= 1.645*0.0209

= 0.0344

4.
=> The 90% confidence interval is (p - E , p + E)

= (0.5595 - 0.0344 , 0.5595 + 0.0344)

= (0.5251 , 0.5939)

5.
=> for 95% confidence interval, Z = 1.960

=> margin of error E = 0.031

=> Sample size n = p*(1 - p)*(Z/E)^2

= 0.5595*(1 - 0.5595)*(1.96/0.031)^2

= 985.2235

= 985 (nearest integer)