In a poll 306 of 1020 randomly selected adults aged 18 or ol
Solution
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!

