Question

Translate the excerpt below into a confidence interval for p and approximate the value of the level of​ confidence, c.

In a survey of 8495​ adults,32.7​%said they were taking vitamin E as a supplement. The​ survey's margin of error is plus or minus 1​%.

a) The confidence interval for the proportion p is ( , )

b) The interval is approximately a___% confidence interval.

Answer 1

Given,

No of adults = 8495

32.7 said they taking vitamin E

Margin of error is plus or minus 1%

Confidence interval for the proportion is

= Sample proportion +/- margin of error

Margin of error = confidence coefficient * standard error of p

Confidence coefficient is the critical value of z = c

Margin of error = 1% = 0.01

Sample proportion (p) = 32.7% = 0.327.

Standard error of p= sqrt[ p*(1-p)/n]

= Sqrt [ 0.327 * (1-0.327) / 8495]

= Sqrt [ 0.327 * 0.673/ 8495 ]

= Sqrt [ 0.220071/ 8495]

= Sqrt [ 0.0000259058]

= 0.00508

Therefore

0.01 = c* 0.00508

C = 0.01/ 0.00508

C= 1.96850

Confidence level corresponding to c= 1.968 is

P(-1.96 < Z < 1.96) *100

0.9500 * 100

95%

a) confidence interval for population proportion is

0.327 +/- 0.01

Upper bound = 0.327 + 0.01= 0.337

Lower bound = 0.327 -.0.01 = 0.317

Required confidence interval [ 0.317 , 0.337]

b) The confidence interval is 95%