Question

In a survey a random sample of 150 households were asked to fill out a questionnaire.
The results showed that 35 households owned more than one car. Find a 99% confidence
interval for the proportion of all households with one car. [4]
The results also included the weekly expenditure on food, dollars, of the households.
These were summarized as follows.

n = 150 Σx = 19035 Σx^2 = 4054716
Find a 95% confidence interval for the mean weekly expenditure on food.

Answer 1

150-35=115$\text{ with one car}.$

$1) \overline{p}=\frac{115}{150}\approx 0.76\,\text{ is a point estimate of the population proportion}.\\ \overline{p}n=(0.76)150>5, (1-\overline{p})n=(0.24)150>5. \\ \text{So the binomial distribution can be estimated by the normal}.\\ p=\overline{p}\pm [z_{\alpha/2}\sqrt{\frac{\overline{p}(1-\overline{p})}{n}}].\\ \alpha=0.01\\ z_{0.005}=2.58\\ p=0.76\pm [(2.58)\sqrt{\frac{(0.76)(0.24)}{150}}].\\ (0.67, 0.85)\,\text{ is our confidence interval}.\\ 2) \mu=\overline{\mu}\pm [z_{\alpha/2}\frac{\sigma}{\sqrt{n}}]\\ \overline{\mu}=\frac{19035}{150}=126.9\,\text{ is a point estimate of the population mean}.\\ \alpha=0.05\\ z_{\alpha/2}=1.96\\ \mu=126.9\pm [(1.96)\frac{\sigma}{\sqrt{150}}]\\ \sigma=\sqrt{\frac{\sum x^2}{n}-(\frac{\sum x}{n})^2}=\sqrt{\frac{4054716}{150}-(126.9)^2}.\\ (110.17, 143.63)\,\text{is the confidence interval}.$