Question 1079177: According to a Pew Research Center nationwide telephone survey of adults conducted between March 15 and April 24, 2011, 55% of college graduates said that college education prepared them for a job (Time, May 30, 2011). Suppose this result was true of all college graduates at that time. In a recent sample of 1950 college graduates, 60% said that college education prepared them for a job. Is there significant evidence at a 10% significance level to conclude that the current percentage of all college graduates who will say that college education prepared them for a job is different from 55%? Use both the p-value and the critical-value approaches.
Round your answers for the observed value of z and the critical value of z to two decimal places, and the p-value to four decimal places. Enter the critical values in increasing order.
z observed =
p-value =
Critical values: ? and ?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! According to a Pew Research Center nationwide telephone survey of adults conducted between March 15 and April 24, 2011, 55% of college graduates said that college education prepared them for a job (Time, May 30, 2011). Suppose this result was true of all college graduates at that time. In a recent sample of 1950 college graduates, 60% said that college education prepared them for a job. Is there significant evidence at a 10% significance level to conclude that[ the current percentage of all college graduates who will say that college education prepared them for a job is different from 55%? Use both the p-value and the critical-value approaches.
Round your answers for the observed value of z and the critical value of z to two decimal places, and the p-value to four decimal places. Enter the critical values in increasing order.
Ho: p = 0.55
Ha: p # 0.55
z(0.60) = (0.60-0.55)/sqrt[0.55*0.45/1950] = 4.438
z observed = 4.438
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p-value = 2*P(z > 4.438) = 2*normalcdf(4.438,100) = 0.000009082
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Critical values: = -1.645 and +1.645
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Cheers,
Stan H.
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