Question

a decade - old study found that the proportion, p, of high school seniors who believed that \getting rich\ was an important personal goal was 60%. a researcher decides to test whether or not that percentage still stands. he finds that, among the 225 high school seniors in his random sample, 178 believe that \getting rich\ is an important goal. can he conclude, at the 0.05 level of significance, that the proportion has indeed changed? perform a two - tailed test. then complete the parts below. carry your intermediate computations to three or more decimal places. (if necessary, consult a list of formulas.) (a) state the null hypothesis h0 and the alternative hypothesis h1. h0:□ h1:□ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) □ (d) find the p - value. (round to three or more decimal places.) □ (e) can we conclude that the proportion of high school seniors who believe that \getting rich\ is an important goal has changed? oyes ono

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the proportion $p$ is equal to the old - known proportion. The alternative hypothesis $H_1$ is that the proportion has changed.
$H_0:p = 0.60$
$H_1:p
eq0.60$

Step2: Determine test - statistic type

We are testing a hypothesis about a proportion, so we use the z - test statistic for a proportion. The formula for the z - test statistic for a proportion is $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$, where $\hat{p}=\frac{x}{n}$, $x$ is the number of successes in the sample, and $n$ is the sample size.

Step3: Calculate $\hat{p}$

Given $x = 178$ and $n=225$, $\hat{p}=\frac{178}{225}\approx0.791$

Step4: Calculate the test - statistic

Substitute $\hat{p}=0.791$, $p = 0.60$, and $n = 225$ into the z - test formula.
$z=\frac{0.791 - 0.60}{\sqrt{\frac{0.60\times(1 - 0.60)}{225}}}=\frac{0.191}{\sqrt{\frac{0.60\times0.40}{225}}}=\frac{0.191}{\sqrt{\frac{0.24}{225}}}=\frac{0.191}{\sqrt{0.0010666\cdots}}\approx\frac{0.191}{0.032659}\approx5.85$

Step5: Calculate the p - value

Since this is a two - tailed test, the p - value is $2\times(1 - P(Z<|z|))$. Using a standard normal table or calculator, for $z = 5.85$, $P(Z < 5.85)\approx1$. So the p - value is $2\times(1 - 1)=0.000$ (to three decimal places).

Step6: Make a decision

Since the p - value ($0.000$) is less than the significance level $\alpha=0.05$, we reject the null hypothesis.

Answer:

(a) $H_0:p = 0.60$; $H_1:p
eq0.60$
(b) z - test statistic for a proportion
(c) $5.850$
(d) $0.000$
(e) Yes