It is believed that 28% 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 2 percentage points?

Answer:Given: The Population Proportion $ (p) = 0.28 $The Margin Of Error $ (E) = 0.02 $The Confidence Interval Level $ = 90\% $ Solution:The Significance Level $ (\alpha) : $$ \alpha = 1- 0.90 $$ = 0.10 $ The…

It is believed that 22% 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 95% confidence and within 3 percentage points?

Answer:Given: The Population Proportion $ (p) = 0.22 $The Margin Of Error $ (E) = 0.03 $The Confidence Interval Level $ = 95% $ Solution:The Significance Level $ (\alpha) : $$ \alpha = 1- 0.95 $$ = 0.05 $ The…

It is believed that 18% 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 4 percentage points?

Answer:Given: The Population Proportion $ (p) = 0.18 $The Margin Of Error $ (E) = 0.04 $The Confidence Interval Level $ = 90\% $ Solution:The Significance Level $ (\alpha) : $$ \alpha = 1- 0.90 $$ = 0.10 $ The…

It is believed that 50% 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 95% confidence and within 2 percentage points?

Answer:Given: The Population Proportion $ (p) = 0.50 $The Margin Of Error $ (E) = 0.02 $The Confidence Interval Level $ = 95\% $ Solution:The Significance Level $ (\alpha) : $$ \alpha = 1- 0.95 $$ = 0.05 $ The…

It is believed that 45% 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) = 0.45 $The Margin Of Error $ (E) = 0.03 $The Confidence Interval Level $ = 90\% $ Solution:The Significance Level $ (\alpha) : $$ \alpha = 1- 0.90 $$ = 0.10 $ The…

It is believed that 40% 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 4 percentage points?

Answer:Given: The Population Proportion $ (p) = 0.40 $The Margin Of Error $ (E) = 0.04 $The Confidence Interval Level $ = 90% $ Solution:The Significance Level $ (\alpha) : $$ \alpha = 1- 0.90 $$ = 0.10 $ The…

It is believed that 35% 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 95% confidence and within 3 percentage points?

Answer:Given: The Population Proportion $ (p) = 0.35 $The Margin Of Error $ (E) = 0.03 $The Confidence Interval Level $ = 95\% $ Solution:The Significance Level $ (\alpha) : $$ \alpha = 1- 0.95 $$ = 0.05 $ The…

It is believed that 20% 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 2 percentage points?

Answer:Given: The Population Proportion $ (p) = 0.20 $The Margin Of Error $ (E) = 0.02 $The Confidence Interval Level $ = 90\% $ Solution:The Significance Level $ (\alpha) : $$ \alpha = 1- 0.90 $$ = 0.10 $ The…

It is believed that 30% 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 95% confidence and within 4 percentage points?

Answer:Given: The Population Proportion $ (p) = 0.30 $The Margin Of Error $ (E) = 0.04 $The Confidence Interval Level $ = 95\% $ Solution:The Significance Level $ (\alpha) : $$ \alpha = 1- 0.95 $$ = 0.05 $ The…