Question

In: Statistics and Probability

A random sample found that forty percent of 50 Americans were satisfied with the gun control...

A random sample found that forty 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 satisfied 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:

Answer= (___ ,___ ) (Keep 3 decimal places)

Expert Solution

SOLUTION:

From given data,

A random sample found that forty 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 satisfied with the gun control laws in 2017. Fill in the blanks appropriately.

Sample proportion = = 40% = 40/100 = 0.40

= 50

97% confidence interval

97% = 97/100 = 0.97

= 1-0.97 = 0.03

/2 = 0.03/2 = 0.015

Critical value:

z_/2 = z_0.015 = 2.1701

A 97% for the true proportion of Americans who were satisfied with the gun control laws in 2017 is:

97% C.I = z_/2 * sqrt(*(1-) / )

97% C.I = 0.40 2.1701* sqrt(0.40*(1-0.40) / 50)

97% C.I = 0.40 0.1503489

97% C.I =( 0.40 -0.1503489 , 0.40 +0.1503489)

97% C.I =(0.249 , 0.550)

orchestra answered 2 years ago

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