Step 1
From the provided information,
Sample size \(\displaystyle{\left({n}\right)}={178}\)
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:
\(\displaystyle{C}{I}=\overline{{{x}}}\pm{z}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

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

\(\displaystyle={2.3}\pm{0.043}\)

\(\displaystyle={\left({2.257},{2.343}\right)}\) 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: \(\displaystyle{C}{I}=\overline{{{x}}}\pm{z}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

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

\(\displaystyle={2.1}\pm{0.012}\)

\(\displaystyle={\left({2.088},{2.112}\right)}\) 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: \(\displaystyle{C}{I}=\overline{{{x}}}\pm{z}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

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

\(\displaystyle={0.78}\pm{0.068}\)

\(\displaystyle={\left({0.712},{0.848}\right)}\) 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: \(\displaystyle{C}{I}=\overline{{{x}}}\pm{z}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

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

\(\displaystyle={6.0}\pm{0.371}\)

\(\displaystyle={\left({5.629},{6.371}\right)}\) Thus, the required confidence interval of viewing television of household is (5.629, 6.371)

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

\(\displaystyle={2.3}\pm{0.043}\)

\(\displaystyle={\left({2.257},{2.343}\right)}\) 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: \(\displaystyle{C}{I}=\overline{{{x}}}\pm{z}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

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

\(\displaystyle={2.1}\pm{0.012}\)

\(\displaystyle={\left({2.088},{2.112}\right)}\) 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: \(\displaystyle{C}{I}=\overline{{{x}}}\pm{z}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

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

\(\displaystyle={0.78}\pm{0.068}\)

\(\displaystyle={\left({0.712},{0.848}\right)}\) 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: \(\displaystyle{C}{I}=\overline{{{x}}}\pm{z}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

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

\(\displaystyle={6.0}\pm{0.371}\)

\(\displaystyle={\left({5.629},{6.371}\right)}\) Thus, the required confidence interval of viewing television of household is (5.629, 6.371)