Please provide complete and detail solutions. Please do it by hand not by comput

Question

Please provide complete and detail solutions. Please do it by hand not by computer software. Thank you!

A survey of 1275 residents of Isla Vista asked whether or not they wanted to build a new train station. 408 of the residents said that they did want to build the station, but 375 said that they did not. 345 people said they had no opinion and 147 refused to respond. Calculate a 95% confidence interval for the percentage who wants to build the station. Calculate a 95% confidence interval for the difference between the proportion who want to build the station and the proportion who do not want to build the station. Estimate the standard error using the sample proportions. Use a chi^2 test to test the null hypothesis that the proportion of the residents that want to build the station is the same as the proportion of the residents that do not.

Explanation / Answer

a.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=408
Sample Size(n)=1275
Sample proportion = x/n =0.32
Confidence Interval = [ 0.32 ±Z a/2 ( Sqrt ( 0.32*0.68) /1275)]
= [ 0.32 - 1.96* Sqrt(0) , 0.32 + 1.96* Sqrt(0) ]
= [ 0.294,0.346]
b.
Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=408
No.Of Observed (n1)=1275
P1= X1/n1=0.32
Proportion 2
No. of chances(X2)=375
No.Of Observed (n2)=1275
P2= X2/n2=0.2941
C.I = (0.32-0.2941) ±Z a/2 * Sqrt( (0.32*0.68/1275) + (0.2941*0.7059/1275) )
=(0.32-0.2941) ± 1.96* Sqrt(0.0003)
=0.0259-0.0358,0.0259+0.0358
=[-0.0099,0.0617]

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