According to the Nielsen Company, the mean number of TV sets in a U.S. household in 2014 was 2.24. Assume the stamdard deviation is 1.1. A sample og 80.

Answer 1

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NOW - I will answer a related question. Two confidence intervals. One for sample, one for sample mean.

Mean TV's in household (µ_TV) = 2.24 Std Dev of sample (σ_TV) = 1.1

If we sampled 80 households at random what is the expected range of TV's we would find in each household? (with confidence of 90%)

(rounding because individual houses can not have fractional TVs)

Range = (2.24- 1.645*1.1) to (2.24- 1.645*1.1) or ( 0 to 4)

If we sampled 80 households at random what is the expected range of the MEAN of the number of tvs per house IN THE SAMPLE? (with 90% confidence)

We calculate the standard deviation of the MEAN of the sample (standard error) = 1.1 / √80 = .123

Range = (2.24- 1.645*.0123) to (2.24- 1.645*.123) or (2.04 to 2.44)

The learning point here is that if we are examining the range of values of what we will find in EACH house, we are looking at population CI we must use the standard deviation in the formula.

If we are looking at what the mean value of a sample is, we are looking at population mean CI and must use standard deviation of mean.