A researcher has gathered information from a random sample of 178 households. For each of the following variables, construct confidence intervals to e

Question

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)

Answer 1

Step 1 From the provided information, Sample size

(n) = 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:

C I = \overset{―}{x} \pm z \frac{s}{\sqrt{n}}

= 2.3 \pm (1.65) \frac{0.35}{\sqrt{178}}

= 2.3 \pm 0.043

= (2.257, 2.343)

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:

C I = \overset{―}{x} \pm z \frac{s}{\sqrt{n}}

= 2.1 \pm (1.65) \frac{0.10}{\sqrt{178}}

= 2.1 \pm 0.012

= (2.088, 2.112)

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:

C I = \overset{―}{x} \pm z \frac{s}{\sqrt{n}}

= 0.78 \pm (1.65) \frac{0.55}{\sqrt{178}}

= 0.78 \pm 0.068

= (0.712, 0.848)

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:

C I = \overset{―}{x} \pm z \frac{s}{\sqrt{n}}

= 6.0 \pm (1.65) \frac{3.0}{\sqrt{178}}

= 6.0 \pm 0.371

= (5.629, 6.371)

Thus, the required confidence interval of viewing television of household is (5.629, 6.371)

A researcher has gathered information from a random sample of 178 households. For each of the following variables, construct confidence intervals to e

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