Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with ? = 2.2%. A random sample of 10 bank stocks gave the following yields (in percents). 5.7 4.8 6.0 4.9 4.0 3.4 6.5 7.1 5.3 6.1 The sample mean is = 5.38%. Suppose that for the entire stock market, the mean dividend yield is ? = 4.4%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.4%? Use ? = 0.01. (a) What is the level of significance? (Enter a number.) State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? H0: ? = 4.4%; H1: ? < 4.4%; left-tailedH0: ? = 4.4%; H1: ? > 4.4%; right-tailed H0: ? > 4.4%; H1: ? = 4.4%; right-tailedH0: ? = 4.4%; H1: ? ≠ 4.4%; two-tailed (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. The standard normal, since we assume that x has a normal distribution with unknown ?.The standard normal, since we assume that x has a normal distribution with known ?. The Student's t, since we assume that x has a normal distribution with known ?.The Student's t, since n is large with unknown ?. Compute the z value of the sample test statistic. (Enter a number. Round your answer to two decimal places.) (c) Find (or estimate) the P-value. (Enter a number. 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) State your conclusion in the context of the application. There is sufficient evidence at the 0.01 level to conclude that the average yield for bank stocks is higher than that of the entire stock market.There is insufficient evidence at the 0.01 level to conclude that the average yield for bank stocks is higher than that of the entire stock market.
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a
- 5.7
- 4.8
- 6.0
- 4.9
- 4.0
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- 6.5
- 7.1
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The sample
(a)
What is the level of significance? (Enter a number.)State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?
(b)
What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.Compute the z value of the sample test statistic. (Enter a number. Round your answer to two decimal places.)
(c)
Find (or estimate) the P-value. (Enter a number. 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 ??(e)
State your conclusion in the context of the application.Trending now
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