Ii the p value cant be statistically significant

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II. The p-value can’t be statistically significant because 0.85 is a considerably high number (greater than the typical alpha level of 0.05). III. You can’t have a 100% CI because the probability that a point estimate equals the true parameter value is zero. IV. A 95% CI for an unknown proportion p is plus or minus two standard errors. V. You must use z* critical value when doing a significance test for a population proportion. Question 2. – Polling Investors a) 1) SE() = = 0.018 2) SE() = = 0.019 b) 95% CI: 1) 0.65 ± 1.96 = (0.615, 0.686) 2) 0.48 ± 1.96 = (0.443, 0.517) c)
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1) There is a 95% probability that the true proportion of investors that have at least $10,000 invested in the stock market is captured within the interval (0.615, 0.686). 2) There is a 95% probability that the true proportion of investors that would be more likely to invest in stocks (over bonds) is captured within the interval (0.443, 0.517). d) 1) = 971.071 = or 972 investors 2) = 1065.404 = or 1066 investors Question 3 – Metal production a) H o : p = p o H a : p < p o p o = 0.25 I will use a one-sided alternative because we are only interested in knowing whether or not the true proportion has decreased from 25% (it can only be smaller), and not equal to, which could be either larger or smaller on both sides. b) Z = = -0.2309 P-value = Pr (Z < -0.23) = 0.4090 Assuming a significance level of 0.05, 0.4090 > alpha, therefore we do not have significant evidence to show that the cracking rate has decreased from 25%. The sample data are compatible
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