Question

In a simple random sample of 190 households, the sample mean number of personal computers was 1.38. Assume the population standard deviation is 0.42.

Construct an 80% confidence interval for the mean number of personal computers. Round the answer to at least two decimal places.

An 80% confidence interval for the mean number of personal computers is
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Answer 1

An 80% confidence interval for the mean number of personal computers in a simple random sample of 190 households with a sample mean of 1.38 and population standard deviation of 0.42 is 1.34 < μ < 1.42.

Step-by-step explanation:

To construct an 80% confidence interval for the mean number of personal computers using a simple random sample, we use the formula:

Confidence interval = sample mean ± (Z-score × (population standard deviation / sqrt(sample size)))

Given that the sample mean is 1.38, population standard deviation is 0.42, and sample size is 190, we need the Z-score for an 80% confidence level. Looking up the Z-score for 80% confidence level in statistical tables, we find it to be approximately 1.28.

Now we can calculate the margin of error (ME):

ME = 1.28 × (0.42 / sqrt(190))

ME = 1.28 × (0.42 / 13.784)

ME = 1.28 × 0.030466

ME = 0.039

(rounded to two decimal places)

The confidence interval is therefore:

1.38 ± 0.039

Lower limit = 1.38 - 0.039 = 1.34

Upper limit = 1.38 + 0.039 = 1.42

So the 80% confidence interval for the mean number of personal computers is 1.34 < μ < 1.42.