Question

An organization monitors many aspects of elementary and secondary education nationwide. Their 1999 numbers are often used as a baseline to assess changes. In 1999 15 % of students had not been absent from school even once during the previous month. In the 2003 ?survey, responses from 8340 randomly selected students showed that this figure had slipped to 14 %. Officials would note any change in the rate of student attendance. Answer the questions below. ?

(a) Write appropriate hypotheses.

Upper H 0 : The percentage of students in 2003 with perfect attendance the previous month is (______)

Upper H Subscript Upper A Baseline : The percentage of students in 2003 with perfect attendance the previous month is (-______) 15%.

?(b) Check the necessary assumptions and conditions.

The independence assumption is(____)

The randomization condition is(____)

The? 10% condition is (___)

The? success/failure condition is(____0

?(c) Perform the test and find the? P-value. ?P-valueequal (-------) ?(Round to three decimal places as? needed.)

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.15
Alternative hypothesis: P 0.15

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it is too small.

b) All conditions are met

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.00391
z = (p - P) /S.D

z = - 2.56

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -2.56 or greater than 2.56.

Thus, the P-value = 0.01

Interpret results. Since the P-value (0.01) is less than the significance level (0.05), we have to reject the null hypothesis.