[5K5) On January 7, 2000, the Gallup Organization released the results of a poll

Question

[5K5) On January 7, 2000, the Gallup Organization released the results of a poll comparin, today with that yesteryear. Then poll results were based in telephone interviews na national sample of 1.031 adults, 18 years and older, conducted December 20-an if the respondent had vacationed for six days or longer within the last 12 months. Supposde attempt to use the poll's results to justify the claim that more than 40 percent o for six days or longer within the last 12 months. The poll actual had done so. Using a confidence interval estimate or a hypothesis test, woul 40 percent of U.S. adults have vacationed for six days or longer within your conclusion tion asked 1999. One ques f U.S. adults have vacationed lly found that 42 percent of the respondents d you conclude that more than the last 12 months? Substantiate I6(5) In the case of Casteneda v. Partida, 430 U.S. 482 (1977), it was found that during a period of 11 years in Hilda County, Texas, 870 people were selected for grand jury duty, and 39% of them were Americans of Mexican ancestry. Among the people eligible for grand jury duty, 79.1% were Americans of Mexican ancestry. There was a claim that the selection process was biased against Americans of Mexican ancestry This case went all the way up to the United States Supreme Court (argued on November 9, 1976, and decided on March 23, 1977). Does the jury selection system appear to be fair? Explain your conclusion in words.

Explanation / Answer

Solution:-

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

Null hypothesis: P < 0.40
Alternative hypothesis: P > 0.40

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 0.05. 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 ]

= 0.0153
z = (p - P) /

z = 1.31

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 greater than 1.31. We use the Normal Distribution Calculator to find P(z < 1.31).

Thus, the P-value = 0.0951

Interpret results. Since the P-value (0.0951) is greater than the significance level (0.05), we have to accept the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that more than 40% of the U.S aduts have vaccinated for six days or longer.

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