A sample of 166 students shows that 65 are part time.... continuted below Give t
ID: 3395059 • Letter: A
Question
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.Explanation / Answer
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.