Question

simple random sample of 25 observations is derived from a normally distributed population with a known standard deviation of 8.2. Use Table 1. a. Is the condition that formula76.mml is normally distributed satisfied? Yes No b. Compute the margin of error with 80% confidence. (Round intermediate calculations to 4 decimal places, "z-value" and final answer to 2 decimal places.) Margin of error 2.0992 c. Compute the margin of error with 90% confidence.(Round intermediate calculations to 4 decimal places, "z-value" and final answer to 2 decimal places.) Margin of error d. Which of the two margins of error will lead to a wider interval? The margin of error with 90% confidence. The margin of error with 80% confidence.

Answer 1

The margin of error with the 80% confidence intervals are 2.099.

80% confidence interval, margin of error can be calculate as follows:

margin of Error :

The margin of Error is a statistic that indicates how much random sampling error is included in the survey results. The greater the margin of error, the lower the confidence that the survey results accurately reflect the census results.

so,n=25, σ=8.2

For 80% confidence intervals, the critical value for z is 1.28. So the required error bound is

ME= Zc.  σ  =1.28 X 8.2 =2.0992

           √n             √25

Therefore, the error bar at 80% confidence is 2.099.

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