Question

An article in Fortune (September 21, 1992) claimed that nearly one-half of all engineers 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. Are the data from Engineering Horizons consistent with the claim reported by Fortune? Use α = 0.05 in reaching your conclusions. Find the P-value for this test
b. Discuss how you could have answered the question in part (a) by constructing a two-sided confidence interval on p.

Answer 1

To determine if the data from Engineering Horizons is consistent with the claim reported by Fortune, we can conduct a hypothesis test using a two-proportion z-test. We compare the test statistic to the critical value to reach a conclusion. Alternatively, we can construct a two-sided confidence interval on the proportion and check if the claim falls within the interval.

Step-by-step explanation:

a. To determine if the data from Engineering Horizons is consistent with the claim reported by Fortune, we can conduct a hypothesis test. The null hypothesis (H0) assumes that the proportion of engineering graduates planning graduate study is equal to one-half. The alternative hypothesis (Ha) assumes that the proportion is different from one-half. We can use a two-proportion z-test to test these hypotheses. The test statistic is calculated as:

Z = (p1 - p2) / sqrt(p * (1-p) * (1/n1 + 1/n2))

Where p1 and p2 are the sample proportions, p is the pooled proportion, and n1 and n2 are the sample sizes. From the given data, p1 = 117/484 = 0.2417, n1 = 484, p2 = 0.50 (proportion from Fortune), n2 = total number of graduates. We can calculate the test statistic and compare it to the critical value for a two-tailed test at α = 0.05. If the test statistic falls within the critical region, we reject the null hypothesis and conclude that the data is not consistent with the claim reported by Fortune. If it falls outside the critical region, we fail to reject the null hypothesis.

b. To construct a two-sided confidence interval on p, we can use the formula:

p ± z * sqrt(p * (1-p) / n)

Where p is the sample proportion, n is the sample size, and z is the critical value based on the desired level of confidence. We can choose a level of confidence (e.g. 95%) and calculate the lower and upper bounds of the confidence interval. If the claim reported by Fortune falls within the confidence interval, we can conclude that the data from Engineering Horizons is consistent with the claim. If the claim falls outside the confidence interval, we can conclude that the data is not consistent with the claim.

Answer 2

a) Null hypothesis:
`p=0.5`

Alternative hypothesis:
`p \\eq 0.5`

`z=\frac{0.242-0.5}{\sqrt{(0.5(1-0.5))/(484)}}=-11.352`

`p_v =2*P(z<-11.352) \approx 0`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of new engineering graduates were planning graduate study differs from 0.5.

b) If we find the confidence interval for the real population proportion and we don't got the 0.5 in the interval we can reject the null hypothesis that p =0.5

`0.242 - 1.96 \sqrt{(0.242(1-0.242))/(484)}=0.204`

`0.242 + 1.96 \sqrt{(0.242(1-0.242))/(484)}=0.280`

And the 95% confidence interval would be given (0.204;0.280). So we see that the upper bound for the confidence interval is <0.5 so we can reject the nu;l hypothesis that p=0.5 at 5% of significance.

Step-by-step explanation:

Data given and notation

n=484 represent the random sample taken

X=117 represent the new engineering graduates were planning graduate study.

`\hat p=(117)/(484)=0.242` estimated proportion of new engineering graduates were planning graduate study.

`p_o=0.5` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of new engineering graduates were planning graduate study is 0.5 or no:

Null hypothesis:
`p=0.5`

Alternative hypothesis:
`p \\eq 0.5`

When we conduct a proportion test we need to use the z statistic, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.242-0.5}{\sqrt{(0.5(1-0.5))/(484)}}=-11.352`

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided
`\alpha=0.05`. The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

`p_v =2*P(z<-11.352) \approx 0`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of new engineering graduates were planning graduate study differs from 0.5.

Part b

If we find the confidence interval for the real population proportion and we don't got the 0.5 in the interval we can reject the null hypothesis that p =0.5

The confidence interval would be given by this formula

`\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}`

For the 95% confidence interval the value of
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.96`

And replacing into the confidence interval formula we got:

`0.242 - 1.96 \sqrt{(0.242(1-0.242))/(484)}=0.204`

`0.242 + 1.96 \sqrt{(0.242(1-0.242))/(484)}=0.280`

And the 95% confidence interval would be given (0.204;0.280). So we see that the upper bound for the confidence interval is <0.5 so we can reject the nu;l hypothesis that p=0.5 at 5% of significance.