Amay 2000 gallup poll found that 38% of a random sample of 1023 adults said that they believe in ghosts. the standard deviation is 1.5%. what is the margin of error for this survey? determine an interval that is likely to contain the percent of all adults who believe in ghosts.

Margin of error = 0.09

Confidence interval = (0.29, 0.41)

Step-by-step explanation:

Sample size, n = 1023

p = 38%

   = 0.38

Standard deviation, σ = 1.5

Consider the level of confidence to be 95%

So, corresponding z* value for 95% confidence level = 1.96

Hence, confidence interval = p ± margin of error

                                               = 0.38 ± 0.09

                                               = (0.29, 0.47)

step-by-step explanation:

divide, then simplify your answer.

1. make -15.66 ÷   5.8 a fraction. then divide.

2. simplify.  

3. final answer.  

2425

step-by-step explanation: