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the national center for education statistics monitors many aspects of e…

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.
a) write appropriate hypotheses.
h₀ : p = 34%
hₐ : p ≠ 34%
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.
the test statistic is z = \square.
(round to two decimal places as needed.)

Explanation:

Step1: State hypotheses

$H_0: p = 0.34$
$H_A: p
eq 0.34$

Step2: Verify assumptions

Independence: The sample is random, and 8705 is less than 10% of all students, so independence holds.
Randomization: The survey uses a random sample, so this is satisfied.
Success/Failure: $np_0 = 8705*0.34 = 2959.7$, $n(1-p_0)=8705*0.66=5745.3$, both are >10, so this is satisfied.

Step3: Calculate sample proportion

$\hat{p} = 0.33$

Step4: Compute standard error

$SE = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.34*0.66}{8705}} \approx \sqrt{\frac{0.2244}{8705}} \approx 0.00508$

Step5: Calculate z-test statistic

$z = \frac{\hat{p}-p_0}{SE} = \frac{0.33-0.34}{0.00508} \approx -1.97$

Step6: Find P-value

For two-tailed test, $P = 2*P(Z < -1.97) = 2*0.0244 = 0.0488$

Answer:

a) $H_0: p = 34\%$, $H_A: p
eq 34\%$
b) All assumptions/conditions are plausibly satisfied as verified.
c) Test statistic: $z = -1.97$, P-value = 0.05 (rounded to two decimal places)
At a typical significance level of $\alpha=0.05$, there is sufficient evidence to conclude the proportion of students absent at least once has changed from 34%.