Question

Suppose a simple random sample of 300 was taken from a large population. The sample had 90% successes. For a 95% confidence level, find the following, rounded to the 4th decimal place. Use your T-Table to find the margin of error. a) How many successes were there? b) - c) d) Margin of Error = Points possible: 4 This is attempt 1 of 3. Message instructor about this question Submit

Answer 1

solution:

Given data

Total No.of samples (n) =300

Proportion of success (p̂) = 90% 0.9

  $\therefore$ Proportion of failure (q̂ ) =1 - p̂ = 1- 0.9 = 0.1

No.of success = 90% of 300 = 0.9 * 300 = 270

To calculate Margin of error,we have

Here Critical value =( 1-c) = 1-0.95 = 0.05

1 -  $\alpha/2$ = 1 - 0.025 = 0.975

From T table for 95% confidence interval ,we have

t* = 1.960

  $\therefore$ Margin of error = t * $\sqrt$ p̂(1-p̂) / n

= 1.960 * $\sqrt$ (0.9*0.1)/300

= 1.960 * 0.0173

= 0.0339

a) No.of Success = 270

b) (p̂) = 0.9

c)  q̂ = 0.1

d) Margin of error (ME) = 0.0339