It is believed that 25% of U.S. homes have a direct satellite television receiver. How large a sample is necessary to estimate the true population of homes that have a direct satellite television receiver with 90% confidence and within 3 percentage points?

Answer:Given: The Population Proportion $ (p) = 0.25 $The Margin Of Error $ (E) = 0.03 $The Confidence Interval Level $ = 90% $ Solution:The Significance Level $ (\alpha) : $$ \alpha = 1- 0.90 $$ = 0.10 $ The…

In a survey, the planning value for the population proportion is p^* = 0.10. How large a sample should be taken to provide a 91% confidence interval with a margin of error of 0.02? Round your answer up to the next whole number.

Answer:Given:The population proportion $(p^*)=0.10$The margin of error $(e)=0.02$The confidence interval level $=91\%$ Solution:→ The significance level at 91% confidence interval:$(\alpha)=1-0.91=0.09$ → The critical value at 0.09 significance level $(z_c)=1.70$ $ \Rightarrow $ The sample size $(n):$→ Here we need to…

In a survey, the planning value for the population proportion is p^* = 0.45. How large a sample should be taken to provide a 98% confidence interval with a margin of error of 0.03? Round your answer up to the next whole number.

Answer:Given:The population proportion $(p^*)=0.45$The margin of error $(e)=0.03$The confidence interval level $=98\%$ Solution:→ The significance level at 98% confidence interval:$(\alpha)=1-0.98=0.02$ → The critical value at 0.02 significance level $(z_c)=2.33$ $ \Rightarrow $ The sample size $(n):$→ Here we need to…

In a survey, the planning value for the population proportion is p^* = 0.15. How large a sample should be taken to provide a 94% confidence interval with a margin of error of 0.09? Round your answer up to the next whole number.

Answer:Given:The population proportion $(p^*)=0.15$The margin of error $(e)=0.09$The confidence interval level $=94%$ Solution:→ The significance level at 94% confidence interval:$(\alpha)=1-0.94=0.06$ → The critical value at 0.06 significance level $(z_c)=1.88$ $ \Rightarrow $ The sample size $(n):$→ Here we need to…

In a survey, the planning value for the population proportion is p^* = 0.50. How large a sample should be taken to provide a 95% confidence interval with a margin of error of 0.04? Round your answer up to the next whole number.

Answer:Given:The population proportion $(p^*)=0.50$The margin of error $(e)=0.04$The confidence interval level $=95\%$ Solution:→ The significance level at 95% confidence interval:$(\alpha)=1-0.95=0.05$ → The critical value at 0.05 significance level $(z_c)=1.96$ $ \Rightarrow $ The sample size $(n):$→ Here we need to…

In a survey, the planning value for the population proportion is p^* = 0.35. How large a sample should be taken to provide a 92% confidence interval with a margin of error of 0.06? Round your answer up to the next whole number.

Answer:Given:The population proportion $(p^*)=0.35$The margin of error $(e)=0.06$The confidence interval level $=92%$ Solution:→ The significance level at 92% confidence interval:$(\alpha)=1-0.92=0.08$ → The critical value at 0.08 significance level $(z_c)=1.75$ $ \Rightarrow $ The sample size $(n):$→ Here we need to…

In a survey, the planning value for the population proportion is p^* = 0.20. How large a sample should be taken to provide a 97% confidence interval with a margin of error of 0.07? Round your answer up to the next whole number.

Answer:Given:The population proportion $(p^*)=0.20$The margin of error $(e)=0.07$The confidence interval level $=97%$ Solution:→ The significance level at 97% confidence interval:$(\alpha)=1-0.97=0.03$ → The critical value at 0.03 significance level $(z_c)=2.17$ $ \Rightarrow $ The sample size $(n):$→ Here we need to…

In a survey, the planning value for the population proportion is p^* = 0.40. How large a sample should be taken to provide a 90% confidence interval with a margin of error of 0.05? Round your answer up to the next whole number.

Answer:Given:The population proportion $(p^*)=0.40$The margin of error $(e)=0.05$The confidence interval level $=90%$ Solution:→ The significance level at 90% confidence interval:$(\alpha)=1-0.90=0.10$ → The critical value at 0.10 significance level $(z_c)=1.645$ $ \Rightarrow $ The sample size $(n):$→ Here we need to…

A federal report indicated that 26% of children under age 6 live in poverty in Oklahoma, an increase over previous years. How large a sample is needed to estimate the true proportion of children under age 6 living in poverty in Oklahoma within 2% with 99% confidence? Round the intermediate calculations to three decimal places and round up your final answer to the next whole number.

Answer: Given: The population proportion $(p)=0.26$ The margin of error $(e)=0.02$ The confidence interval level $=99%$ Solution: → The critical value at 99% confidence interval = 2.576 → The sample size $(n):$ Here we use a margin of error formula…

The estimate of the population proportion is to be within plus or minus 0.04, with an 84% level of confidence. How large a sample is required? (The final answer must be a whole number.)

Answer: Given Data: The margin of error $(e)=0.04$ The confidence interval level $=84%$ Here the sample proportion is not given so we can use $(p)=0.5$ Solution: The level of significance $(\alpha)=1-0.84=0.16$ The critical value $(z_c)=Z_{\frac{\alpha}{2}} = Z_{\frac{0.16}{2}} = 1.41 $…