A random sample found that thirty-six percent of 50 Americans were satisfied with the gun control laws in 2017. Compute a 97% confidence interval for the true proportion of Americans who were satisifed with the gun control laws in 2017. Fill in the blanks appropriately. A 97% for the true proportion of Americans who were satisfied with the gun control laws in 2017 is(_____)(_____)(Keep 3 decimal places)
= 1 - 0.97 = 0.03
From Z table,
Z/2 = 2.1701 (For two tailed confidence interval)
97% confidence interval for p is
[ - Z/2 * sqrt ( ( 1 - ) / n ] < p < [ + Z/2 * sqrt ( ( 1 - ) / n ]
0.36 - 2.1701 * sqrt [ 0.36 * ( 1 - 0.36) / 50] < p < 0.36 + 2.1701 * sqrt [ 0.36 * ( 1 - 0.36) / 50]
0.213 < p < 0.507
97% CI is ( 0.213 , 0.507 )
A random sample found that thirty-six percent of 50 Americans were satisfied with the gun control...
A random sample found that forty-two percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 92% confidence interval for the true proportion of Americans who were satisifed with the gun control laws in 2017. Fill in the blanks appropriately. A 92% for the true proportion of Americans who were satisfied with the gun control laws in 2017 is(____)(_____)(Keep 3 decimal places)
There is no prior information about the proportion of Americans who support gun control in 2019. If we want to estimate 97% confidence interval for the true proportion of Americans who support gun control in 2019 with a 0.27 margin of error, how many randomly selected Americans must be surveyed? Answer:
There is no prior information about the proportion of Americans who support the gun control in 2018. If we want to estimate 92% confidence interval for the true proportion of Americans who support the gun control in 2018 with a 0.24 margin of error, how many randomly selected Americans must be surveyed?
There is no prior information about the proportion of Americans who support gun control in 2019. If we want to estimate 93% confidence interval for the true proportion of Americans who support gun control in 2019 with a 0.18 margin of error, how many randomly selected Americans must be surveyed?
There is no prior information about the proportion of Americans who support the gun control in 2018. If we want to estimate 95% confidence interval for the true proportion of Americans who support the gun control in 2018 with a 0.2 margin of error, how many randomly selected Americans must be surveyed? (Round up your answer to nearest whole number)
A 2007 Harris poll took a random sample of 1241 Americans and found that 493 of them believed that the profession of teaching "has very great prestige". Find and interpret a 98% confidence interval for the true proportion of Americans who believe that the teaching profession "has very great prestige. Find and interpret a 98% confidence interval for the true proportion of Americans who believe that the teaching profession "has very great prestige."
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