In a poll, 306 of 1020 randomly selected adults aged 18 or older stated that the

Question

In a poll, 306 of 1020 randomly selected adults aged 18 or older stated that they believe there is too little spending on national defense. Use this information to complete parts (a) through (e) below. Obtain a point estimate for the proportion of adults aged 18 or older who feel there is too little spending on national defense, Construct a 95% confidence interval for the proportion of adults aged 18 or older who believe there is too little spending on national defense. The 95% confidence interval is Use the results of part (c) to construct a 95% confidence interval for the population proportion of adults aged 18 or older who do not believe there is too little spending on national defense. The 95% confidence interval is (Round to three decimal places as needed.)

Explanation / Answer

a)

p^ = point estimate of the population proportion = x / n = 306/1020 =   0.3 [ANSWER]

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C)          

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.3          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.014348601          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.028122741          
lower bound = p^ - z(alpha/2) * sp =   0.271877259          
upper bound = p^ + z(alpha/2) * sp =    0.328122741          
              
Thus, the confidence interval is              
              
(   0.271877259   ,   0.328122741   ) [ANSWER]

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b)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.7          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.014348601          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.028122741          
lower bound = p^ - z(alpha/2) * sp =   0.671877259          
upper bound = p^ + z(alpha/2) * sp =    0.728122741          
              
Thus, the confidence interval is              
              
(   0.671877259   ,   0.728122741   ) [ANSWER]

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You can use InvNorm(0.025,0,1) to get z(alpha/2) = 1.959963985 here, if you have a ti-84. Thanks!

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