Question

A recent article in USA Today reported that a job awaits 33% of new college graduates. The major reasons given were an overabundance of college graduates and a weak economy. A survey of 200 recent graduates from your school revealed that 80 students had jobs. At a 99% level of confidence, can we conclude that a larger proportion of students at your school have jobs?

(a) State the null and alternate hypothesis. (b) Determine which distribution to use for the test statistic, and state the level of significance. (c) Calculate the necessary sample test statistics (d) Draw a conclusion and interpret the decision.

Answer 1

(a)

Let p be the proportion of students at the school have jobs.

Null hypothesis H0: p = 0.33

Alternative hypothesis Ha: p > 0.33

(b)

Below conditions for one-sample z test are satisfied.

• The sampling method is simple random sampling.
• Each sample point can result in just two possible outcomes. We call one of these outcomes a success for students have jobs and the other, a failure for students who do not have jobs .
• The sample includes 80 students had jobs and 120 students do not had jobs and thus at least 10 successes and 10 failures.
• Obviously, the population size of school students is at least 20 times as big as the sample size of 200.

The distribution of the sample proportion is normal distribution with mean as = 80/200 = 0.4 and standard deviation as = 0.03464

The test statistic is z statistic.

For this analysis, the significance level is 0.05.

(c)

Test statistic, z = (0.4 - 0.33) / 0.03464 = 2.020785

(d)

P-value = P(z > 2.02) = 0.0217

As, p-value is less than the significance level of 0.05, we reject the null hypothesis H0 and conclude that there is significant evidence that a larger proportion of students at your school have jobs.