Last year a bill was passed. Of those who voted, 40 people gave two pieces of information: if they voted yes or no and their height (in inches). Assuming the samples were drawn from populations with the same population variance, find the 95% confidence interval for the difference in means of heights.
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Last year a bill was passed. Of those who voted, 40 people gave two pieces of information: if they voted yes or no and their height (in inches).
Assuming the samples were drawn from populations with the same population variance, find the 95% confidence interval for the difference in means of heights.
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- Conditional probability If 40 of the population have completed college, and 85 of college graduates are registered to vote, what percent of the population are both college graduates and registered voters?Levene's test of equality variances shows: 1. The result is significant, so the variances are equal 2. The result is significant, so the variances are unequal 3. The result is nonsignificant, so the variances are equal 4. The result is nonsignificant, so the variances are unequal From the Levene's test result, what might you conclude about the use of the t test? 1. Since the variances are equal, the assumption of equal variances is met 2. Since the variances are equal, the assumption of equal variances is not met 3. Since the variances are unequal, the assumption of equal variances is met 4. Since the variances are unequal, the assumption of equal variances is not met.Levene’s test of equality of variances shows: 1. The result is significant, so the variances are equal 2. The result is significant, so the variances are unequal 3. The result is nonsignificant, so the variances are equal 4. The result is nonsignificant, so the variances are unequal From the results, what might you conclude about the use of the t test? 1. Since the variances are equal, the assumption of equal variances is met 2. Since the variances are equal, the assumption of equal variances is not met 3. Since the variances are unequal, the assumption of equal variances is met 4. Since the variances are unequal, the assumption of equal variances is not met
- Setup: JP took two random samples from different populations.Sample 1: A random sample of size n1 = 15 is taken from an approximate normalpopulation. From our sample, we find a variance of s1^2= 11.1672, and a mean of x¯1 = 115.1192.Sample 2: A second random sample of size n2 = 23 is taken from a differentapproximate normal population. From our second sample, we find a variance ofs2^2 = 6.310136, and a mean of ¯x2 = 125.0925.Goal: We wish to test the hypothesis that µ1 − µ2 = −8,against the alternative µ1 − µ2 < −8 at a significance level of 2.9%.R Output:> t.test(data1,data2, mu = -8, var.equal = FALSE,alternative = "less", conf.level = 0.971)Welch Two Sample t-testdata: data1 and data2t = -1.955, df = 24.134, p-value = 0.03113alternative hypothesis: true difference in means is less than -897.1 percent confidence interval:-Inf -7.964374sample estimates:mean of x mean of y115.1192 125.0925Questions:(a) Why was the test run with var.equal = FALSE?(b) State the appropriate p-value,…Setup: JP took two random samples from different populations.Sample 1: A random sample of size n1 = 15 is taken from an approximate normalpopulation. From our sample, we find a variance of s1^2= 11.1672, and a mean of x¯1 = 115.1192.Sample 2: A second random sample of size n2 = 23 is taken from a differentapproximate normal population. From our second sample, we find a variance ofs2^2 = 6.310136, and a mean of ¯x2 = 125.0925.Goal: We wish to test the hypothesis that µ1 − µ2 = −8,against the alternative µ1 − µ2 < −8 at a significance level of 2.9%.R Output:> t.test(data1,data2, mu = -8, var.equal = FALSE,alternative = "less", conf.level = 0.971)Welch Two Sample t-testdata: data1 and data2t = -1.955, df = 24.134, p-value = 0.03113alternative hypothesis: true difference in means is less than -897.1 percent confidence interval:-Inf -7.964374sample estimates:mean of x mean of y115.1192 125.0925 (d) State your conclusion.(e) If you were doing this hypothesis test using critical regions…3. A recent report stated that the proportion of teens that vape has decreasedfrom the prior year. The authors randomly selected a group of 600 teensfrom the current year and 126 of them vape. They randomly selected 800teens the prior year and 216 of them vaped. Is there significant evidencethat the proportion decreased?a. State the hypotheses to be tested.b. Compute the value of the pooled proportion
- Salinity in Quebec river is a combination of tides and rain. Rain that diluted the salt occurs about 10% of the time, while the non salty outgoing tides occurs twice per day. You take your dog for a walk down by the river at random during the weekend and notice that he'll drink from the river 30% of the time when the salt is sufficiently low. Question somebody else points that it's more likely to rain in the afternoon. You happen to be the time of year when half the tides are also occurring in the afternoon ( the others occur roughly 12 hours later during the night) if the probability of rain during this outgoing tide period increases to 15 % what is the probability of there being an afternoon tide during a rain ?Salinity in Quebec river is a combination of tides and rain. Rain that diluted the salt occurs about 10% of the time, while the non salty outgoing tides occurs twice per day. You take your dog for a walk down by the river at random during the weekend and notice that he'll drink from the river 30% of the time when the salt is sufficiently low. Questions somebody else points that it's more likely to rain in the afternoon. You happen to be the time of year when half the tides are also occurring in the afternoon ( the others occur roughly 12 hours later during the night) if the probability of rain during this outgoing tide period increases to 15 % what is the probability of there being an afternoon tide during a rain ? Note : ( the probability of there being a tide and rain together is NOT the same as the probability of there being a tide during a rain) Given that the overall probability of rain remains 10 % what is the probability of rain during a non tide period?A researcher's survey results indicate that about 50% of 12th graders consume less than 300g of vegetables at lunch, about 40% consume 300-600g, and only 10% of them consume more than 600g. A high school conducted a survey of student diets and health information, and reported data about 12th graders' vegetable intake at lunch is reported in the table photo. If researchers wants to find out whether the high school has the same proportions of 12th graders in the vegetable intake categories as reported by the survey, what would the appropiate test statistic be to use and why. Choosing between an ANOVA, Chi -Square independence tests, Chi Square Goodness of Fit, and One sample proportion test?
- Answers: 2. t = .656, do not reject Ho: there is no evidence that this year's students are performing differently than students in previous years. 3. chi-square = 7.997, Do not reject H0; no evidence that the variance is significantly less than 175. chi-square = 11.250, Do not reject H0; no evidence that the variance is significantly greater than 175. 4. 3.973 to 13.5219Economics Now suppose that the time series process {Xt}, is expressed as Xt = z + et where et is iid with a mean of zero and a variance of σ , and the variable z does not change over time (time invariant) which means it has a mean E(z) = 0 and , and it is assumed that z and et are uncorrelated: i. Find the mean xt , E(Xt) and the variance Xt, var(Xt). Do they depend on t? ii. Determine the covariance of xt and xt+h for h > 0, Cov(Xt , xt+h) iii. Is xt stationary? Explain.Consider the game of craps designed by Econ 261 Hotel students. The game consists of rolling two fair six-sided dice. You win a dollar if the sum of the dots on the two dice is 2, 3, 4, or 5; if the sum of the dots on the two dice is 9, 10, 11, or 12 you lose a dollar. You win nothing, (that is you get $0) if the sum is 6, 7, or 8. The variance of X, Var(X) is what?