Chemistry homework help. USA Today reported that about 47% of the general consumer population in the United States is loyal to the automobile manufacturer of their choice. Suppose Chevrolet did a study of a random sample of Chevrolet owners and found that said they would buy another Chevrolet. Does this indicate that the population proportion of consumers loyal to Chevrolet is more than 47%? Use ? = 0.01. (a) What is the level of significance? State the null and alternate hypotheses. H0: p = 0.47; H1: p ? 0.47 H0: p > 0.47; H1: p = 0.47    H0: p = 0.47; H1: p > 0.47 H0: p = 0.47; H1: p < 0.47 (b) What sampling distribution will you use? The Student's t, since np < 5 and nq < 5. The standard normal, since np < 5 and nq < 5.    The standard normal, since np > 5 and nq > 5. The Student’s t, since np > 5 and nq > 5. What is the value of the sample test statistic? (Round your answer to two decimal places.) (c) Find the P-value of the test statistic. (Round your answer to four decimal places.)(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ?? At the ? = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.    At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (e) Interpret your conclusion in the context of the application. There is sufficient evidence at the 0.01 level to conclude that the true proportion of customers loyal to Chevrolet is more than 0.47. There is insufficient evidence at the 0.01 level to conclude that the true proportion of customers loyal to Chevrolet is more than 0.47.

Chemistry homework help