 # A researcher has gathered information from a random sample of 178 households. For each of the following variables, construct confidence intervals to e Lennie Carroll 2020-12-30 Answered
A researcher has gathered information from a random sample of 178 households. For each of the following variables, construct confidence intervals to estimate the population mean. Use the 90% level. a. An average of 2.3 people resides in each household. Standard deviation is 0.35. b. There was an average of 2.1 television sets (s  0.10) and 0.78 telephones (s  0.55) per household. c. The households averaged 6.0 hours of television viewing per day (s  3.0)
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Step 1 From the provided information, Sample size $\left(n\right)=178$ The sample is very large therefore z-distribution will be used to obtain the confidence interval. The z value at 90% confidence level from the standard normal table is 1.65. a) For average of 2.3 with standard deviation 0.35 the required confidence interval can be obtained as: $CI=\stackrel{―}{x}±z\frac{s}{\sqrt{n}}$
$=2.3±\left(1.65\right)\frac{0.35}{\sqrt{178}}$
$=2.3±0.043$
$=\left(2.257,2.343\right)$ Thus, the confidence interval of people resides in each household is (2.257, 2.343) Step 2 b) Televisions sets per households with the average 2.1 and standard deviation 0.10. The required confidence interval for the television set per household can be obtained as: $CI=\stackrel{―}{x}±z\frac{s}{\sqrt{n}}$
$=2.1±\left(1.65\right)\frac{0.10}{\sqrt{178}}$
$=2.1±0.012$
$=\left(2.088,2.112\right)$ Thus, the confidence interval of televisions per household is (2.088, 2.112) The required confidence interval for the telephones set per household can be obtained as: $CI=\stackrel{―}{x}±z\frac{s}{\sqrt{n}}$
$=0.78±\left(1.65\right)\frac{0.55}{\sqrt{178}}$
$=0.78±0.068$
$=\left(0.712,0.848\right)$ Thus, the confidence interval of telephones per household is (0.712, 0.848). Step 3 c) Television viewing per day of each households has an average of 6.0 hours with standard deviation 3.0. The required confidence interval can be obtained as: $CI=\stackrel{―}{x}±z\frac{s}{\sqrt{n}}$
$=6.0±\left(1.65\right)\frac{3.0}{\sqrt{178}}$
$=6.0±0.371$
$=\left(5.629,6.371\right)$ Thus, the required confidence interval of viewing television of household is (5.629, 6.371)