An article in Fortune (September 21, 1992) claimed that nearly one-half of all engineering continue academic studies beyond the B.S. degree, ultimately receiving either an M.S. or a Ph.D. degree. Data from an article in Engineering Horizons (Spring 1990) indicated that 117 of 484 new engineering graduates were planning graduate study.
a) Is the data from Engineering Horizons consistent with the claim reported by Fortune? Use α=0.05 in reaching your conclusions.
b) Find P-Value for this test.
c) Discuss how you could answered the question in part-a by constructing a two-sided confidence interval on p.
I'm not even sure where to start. Any help would be much appreciated. Thank you.
To make sense of this problem, you have to assume that the population from which EH is sampling is the same as the population Fortune is talking about, or at least that you want to check the '1/2 claim' based on some other reason.
Note: As stated in the comment on this site, an issue is whether Fortune means 'eventually' or 'as new graduates'. In ANY poll the question always arises whether we sampled randomly from the intended population. Statistical methods only deal with random sampling error, not with sampling from the wrong population. (Very carefully taking a random sample at Republican Party Headquarters is hardly a good way to predict the outcome of an upcoming election for governor of a US state.)
Now that the philosophical sermon is done, I take the question as asking for help on the mechanics of a particular kind of hypothesis test and confidence interval. I hope the following provides useful guidance.
Provided the EH data are randomly sampled from a population we care about, we might use them to test whether the proportion of "new students planning to continue' in that population is 1/2.
A sample of size $n = 484$ to test $H_0: p = 1/2$ against a two sided alternative (replacing $=$ with $\ne$) lends itself to a normal approximation. Then reject the null hypothesis (and say 'inconsistent') since $|(\hat p - 1/2)/\sqrt{1/4(484)}| = 11.84 > 1.96,$ where $\hat p = 117/ 484.$
Note: P-value is probability of a value under the standard normal curve outside of the interval $(-11.36, 11.36).$ The answer is essentially 0 (to many places; $6.6 \times 10^{−30}$). For practical purposes, almost all of the area under a standard normal curve is between $(-4, 4).$ The number 11.36 is way beyond any value you will find in your normal table.
A traditional 95% confidence interval would be $\hat p \pm 1.96\sqrt{\hat p (1-\hat p)/484}$. This traditional interval may be a reasonable choice for a sample size as large as 484. But (for the 95% level ONLY) some books recommend using a more accurate interval $\hat p^* \pm 1.96\sqrt{\hat p^* (1-\hat p^*)/488}$, where $\hat p^* = (117+2)/(484+4)$.
[For the latter style of confidence interval, I got (0.206 0.282), which does not include 1/2. Thus the EH data are not consistent with half of the new graduates planning to continue. If this is homework, you need to verify that this is the intended formula, do the computation, and explain the result.)