Words 3 points Sa QUESTION 3 It has been found that 85 6% of all enrolled colleg

Question

Words 3 points Sa QUESTION 3 It has been found that 85 6% of all enrolled college and university stdents in the United States are undergraduates A random sample of 500 enrolled students in a particular state revealed that 420 of them were undergraduates. Is there sufficient evidence to conclude that the proportion of differs from the national average? Use aplha-0.01 Define Step 1 Step 2 Step 3 Step 4 Step 5 Pathp Click Seve and Submit to sase and submitClick Saue AllAnsuers to save allanson Save Al Answrers

Explanation / Answer

Solution:-

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

Null hypothesis: P = 0.856
Alternative hypothesis: P 0.856

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.

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 () and compute the z-score test statistic (z).

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

= 0.0157
z = (p - P) /

z = - 1.02

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 -1.02 or greater than 1.02.

Thus, the P-value = 0.308

Interpret results. Since the P-value (0.308) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we have sufficient evidence that the proportion do not differ from the national average.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.