A sample of 166 students shows that 65 are part time continu

A sample of 166 students shows that 65 are part time.... continuted below

Give the following with respect to the question in italics Critical Value with the correct distribution symbol -t or z Give the numeric value of the sample statistic plusminus the numeric value of the margin of error. You will be given a +1 bonus for showing formula input for computation to get the given value. Give the confidence interval as a range of values using a compound inequality with the correct population parameter\'s symbol notated in the middle. Round appropriately with respect to the distribution of the random variable. Interpret the confidence interval. A sample of 166 students shows that 65 are part-time. Construct a 95% confidence interval for the true proportion of students that are part-time. Assume that all assumptions are met for constructing such an interval.

Solution

i)

Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          

ii)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.391566265          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.037883948          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.074251174  
Thus,

0.391566265 - 0.074251174 < p < 0.391566265 + 0.074251174

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iii)
      
lower bound = p^ - z(alpha/2) * sp =   0.317315091          
upper bound = p^ + z(alpha/2) * sp =    0.465817439          
              
Thus, the confidence interval is          
0.317315091 < p < 0.465817439

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iv)

We are 95% confident that the true population proportion of those who work part time is between 0.317315091 and 0.465817439.