Confdence interval for one population proportion Archives

Confdence interval for one population proportion

Microwave Cooking A survey of 115 households found that 69% use the microwave to prepare their meals. Find the 95% confidence interval for the proportion. Use Excel and round the answers to three decimal places.

Answer:Given:The sample proportion $(\hat{p}) = 0.69$ The sample size $(n) = 115$ The confidence interval level $= 95 %$ Solution:The significance level $(α) :$ $α = 1 - 0.95$ $= 0.05$ The…

Confdence interval for one population proportion

Out of a sample of 35 people, 15 individuals have asthma. Compute a 95% confidence interval for the proportion of individuals who have asthma.

Answer:Given:Favorable Cases $(x) = 15$ Sample Size $(n) = 35$ Confidence Interval Level $= 95 %$ Solution:The Sample Proportion $(\hat{p}) :$ $\hat{p} = \frac{15}{35} = 0.4286$ The Significance Level $(α) :$ $α = 1 - 0.95 = 0.05$ The critical value $(z_{c}) :$ $z_{c} = Z_{α / 2} = Z_{0.05 / 2} = 1.96$ The…

Confdence interval for one population proportion

Out of a sample of 45 people, 20 individuals are obese. Compute a 95% confidence interval for the proportion of individuals who are obese.

Answer:Given:Favorable Cases $(x) = 20$ Sample Size $(n) = 45$ Confidence Interval Level $= 95 %$ Solution:The Sample Proportion $(\hat{p}) :$ $\hat{p} = \frac{20}{45} = 0.4444$ The Significance Level $(α) :$ $α = 1 - 0.95 = 0.05$ The critical value $(z_{c}) :$ $z_{c} = Z_{α / 2} = Z_{0.05 / 2} = 1.96$ The…

Confdence interval for one population proportion

Out of a sample of 60 people, 25 individuals are smokers. Compute a 95% confidence interval for the proportion of individuals who are smokers.

Answer:Given:Favorable Cases $(x) = 25$ Sample Size $(n) = 60$ Confidence Interval Level $= 95 %$ Solution:The Sample Proportion $(\hat{p}) :$ $\hat{p} = \frac{25}{60} = 0.4167$ The Significance Level $(α) :$ $α = 1 - 0.95 = 0.05$ The critical value $(z_{c}) :$ $z_{c} = Z_{α / 2} = Z_{0.05 / 2} = 1.96$ The…

Confdence interval for one population proportion

Out of a sample of 30 people, 12 individuals have diabetes. Compute a 95% confidence interval for the proportion of individuals who have diabetes.

Answer:Given:Favorable Cases $(x) = 12$ Sample Size $(n) = 30$ Confidence Interval Level $= 95 %$ Solution:The Sample Proportion $(\hat{p}) :$ $\hat{p} = \frac{12}{30} = 0.4$ The Significance Level $(α) :$ $α = 1 - 0.95 = 0.05$ The critical value $(z_{c}) :$ $z_{c} = Z_{α / 2} = Z_{0.05 / 2} = 1.96$ The…

Confdence interval for one population proportion

Out of a sample of 50 people, 22 individuals have high cholesterol. Compute a 95% confidence interval for the proportion of individuals who have high cholesterol.

Answer:Given:Favorable Cases $(x) = 22$ Sample Size $(n) = 50$ Confidence Interval Level $= 95 %$ Solution:The Sample Proportion $(\hat{p}) :$ $\hat{p} = \frac{22}{50} = 0.44$ The Significance Level $(α) :$ $α = 1 - 0.95 = 0.05$ The critical value $(z_{c}) :$ $z_{c} = Z_{α / 2} = Z_{0.05 / 2} = 1.96$ The…

Confdence interval for one population proportion

Out of a sample of 40 people, 18 individuals have high blood pressure. Compute a 95% confidence interval for the proportion of individuals who have high blood pressure.

Answer:Given:Favorable Cases $(x) = 18$ Sample Size $(n) = 45$ Confidence Interval Level $= 95$ Solution:The Sample Proportion $(p \hat{)} :$ $p \hat{=} \frac{18}{45} = 0.4$ The Significance Level $(α) :$ $α = 1 - 0.95 = 0.05$ The critical value $(z_{c}) :$ $ z_c…