Question

the national center for education statistics monitors many aspects of elementary and secondary education nationwide. their 1996 numbers are often used as a baseline to assess changes. in 1996, 34% of students had not been absent from school even once during the previous school year. in a 2000 survey, responses from 8705 students showed that this figure had slipped to 33%. officials would, of course, be concerned if student attendance were declining. do these figures give evidence of a change in student attendance? complete parts a through e below.
b) check the assumptions and conditions.
the independence assumption is plausibly justified.
the randomization condition is plausibly satisfied.
the 10% condition is plausibly satisfied.
the success/failure condition is satisfied.
c) perform the test and find the p - value.
1.97 (aces as needed)
places as needed)
use α = 0.10.
do not reject
reject
the null hypothesis. there
sufficient evidence to conclude that the proportion of students with perfect attendance

Explanation:

Step1: State hypotheses

Let $p$ = proportion of students with perfect attendance.
$H_0: p = 0.34$
$H_a: p
eq 0.34$

Step2: Calculate sample proportion

Sample proportion $\hat{p} = 0.33$, $n=8705$
$\hat{p} = 0.33$

Step3: Compute test statistic

Use z-test for proportion:
$$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$$
$$z = \frac{0.33 - 0.34}{\sqrt{\frac{0.34(1-0.34)}{8705}}} = \frac{-0.01}{\sqrt{\frac{0.2244}{8705}}} \approx -1.97$$

Step4: Find P-value

Two-tailed test, so P-value = $2 \times P(Z < -1.97)$
From z-table, $P(Z < -1.97) = 0.0244$, so P-value = $2 \times 0.0244 = 0.0488$

Step5: Compare to α

$\alpha = 0.10$, P-value = 0.0488 < 0.10

Answer:

Reject the null hypothesis. There is sufficient evidence to conclude that the proportion of students with perfect attendance has changed.
P-value: 0.0488