Question

For Problems 6-10, a national survey shows that 48.3% of Americans support impeaching the President of the United States while the rest either do not support impeachment or are indifferent to the impeachment process. [Source: https://projects.fivethirtyeight.com/impeachment-polls/; as of November 3, 2019]

6. An independent survey takes a sample of 65 individuals, finding that 31 of them are in favor of impeachment. Find the mean, standard deviation, and standard error of the sample. Be sure to use the correct notation based on the sample size!

7. Find the 99% confidence interval for the sample in Problem 6

Answer 1

(6)

Mean = n * P

= ( 65 * 0.483 )

  = 31.395
Variance = n * P * Q

= ( 65 * 0.483 * 0.517 )

= 16.2312
Standard deviation = √(variance)

= √(16.2312)

= 4.0288

Standard Error = √( (p * q) / n)

Standard Error = 0.0620

( 7)

p̂ = X / n = 31/65

= 0.4769
p̂ ± Z(α/2) √( (p * q) / n)
0.4769 ± Z(0.01/2) √( (0.4769 * 0.5231) / 65)
Z(α/2) = Z(0.01/2) = 2.576
Lower Limit = 0.4769 - Z(0.01) √( (0.4769 * 0.5231) / 65)

= 0.3173
upper Limit = 0.4769 + Z(0.01) √( (0.4769 * 0.5231) / 65)

= 0.6365
99% Confidence interval is ( 0.3173 , 0.6365 )
( 0.3173 < P < 0.6365 )

------------------------------------------------------------------------

DEAR STUDENT,

IF YOU HAVE ANY QUERY ASK ME IN THE COMMENT BOX,I AM HERE TO HELPS YOU.PLEASE GIVE ME POSITIVE RATINGS

*****************THANK YOU***************