number 5 conlmited, expluin een committed, explain what it woold have been prove
ID: 3322845 • Letter: N
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number 5
conlmited, expluin een committed, explain what it woold have been prove? ther states that StopSmoke significanily ey referring to statistical significance? What other type significance. Oue states thbat StopS moke provides significant of significance e is significantly less expensive than other ers Use the traditional method of hypothesis testing unless otherwise specified. 5. Life on Other Planets Forty six percent of people ns. believe that there is life on other planets in the universe. A scientist does not agree with this finding. He surveyed 120 randomly selected individuals and found 48 believed that there is life on other planets. At : 0.10, is there sufficient evidence to conclude that the percentage differs from 46? Soance: American Health, Inc rng 6. Stocks and Mutual Fund Ownership It has been found that 50.3% of U.S. households own stocks and mutual funds. A random sample of 300 heads of households indicated that 171 owned some type of stock. At what level of significance would you conclude that this was a significant difference? Source: www.census,gov 8-45Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P = 0.46
Alternative hypothesis: P 0.46
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 0.10. 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 ]
= 0.0455
z = (p - P) /
z = - 1.32
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 -1.32 or greater than 1.32. Thus, the P-value = 0.1868
Interpret results. Since the P-value (0.1868) is greater than the significance level (0.10), we cannot reject the null hypothesis.
We do not have sufficient evidence to conclude that propoortion differs from 0.46.
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