Question

Example: The 95% CI of the mean weight (in pounds) of apples sold as oversized is (0.54, 0.98). This CI was based on a random sample of 44 apples. What is the margin of error of this estimate?

a)0.22

2. b) 0.44

3. c) 0.0325

4. d) 0.76

5. e) not enough information

If the ME is to be reduced to 0.0152, while 95% confidence is to be maintained, approximately what size should the sample be? Assume  = 0.1.

1. a) 56

2. b) 167

3. c) 106

4. d) 198

5. e) None of the above.

Answer 1

Q.1) Given that, the 95% CI of the mean weight of apples sold as oversized is (0.54, 0.98)

where, lower limit = 0.54 and upper limit = 0.98

The margin of error ( ME ) of estimate is,

ME = (upper limit - lower limit)/2 = (0.98 - 0.54)/2 = 0.44/2 = 0.22

The margin of error of this estimate is 0.22

Q.2) Given that, margin of error (ME) = 0.0152

standard deviation = 0.1

A 95% confidence level has significance level = 0.05 and critical value is,

We want to find, the sample size ( n ),

n = 167 (rounded to next whole number)

Required sample size is 167