All answers must include I. the rationale of using a test, 2. the hypotheses, 3.

Question

All answers must include I. the rationale of using a test, 2. the hypotheses, 3. the critical-value, 4. the test-statistics, 5. the p-value, 6. the test result, 7. the distribution chart, and 8. the conclusion I. In baseball, league A allows a designated hitter (DHI to bat for the pitcher. In league B, the pitcher must bat. The common belief is that this results in league A teams scoring more runs n interleague play, when league A teams visit league B teams, the league A pitcher must bat. So, if the DH does result in more runs, it would be expected that league A teams will score fewer runs when visiting league B parks. To test this claim, a random sample of runs scored by league A teams with and without their DH is as the followings: n': 60, 60 , s4-35, n,-50, 4.2, and SB-2. Does the designated hitter result in more runs scored at the 0.0 7level ofsigni ficance? . 6. 7. 2. An organization surveyed 1040 adults in a certain country aged 18 and older and found that 537 believed they would not have enough money to live comfortably in retirement. Does the sample evidence suggest that a majority of adults in this country believe they will not have enough money in retirement? Use the = 0.09 level ofsignificance. 6. 2. 7

Explanation / Answer

Solution:-

2)

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

Null hypothesis: P < 0.50
Alternative hypothesis: P > 0.50

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.09. 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 ]

= 0.0155
z = (p - P) /

z = 3.29

zcritical = 1.7

Rejection region z > 1.7

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 one-tailed test, the P-value is the probability that the z-score is more than 3.29.

Thus, the P-value = 0.0005

Interpret results. Since the P-value (0.0005) is less than the significance level (0.09), we cannot accept the null hypothesis.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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