QUESTION IMAGE
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.)
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, $P = 2*P(Z < -1.97) = 2*0.0244 = 0.0488$
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
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%.