Final answer:
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.