Give a 99.9% confidence interval, for μ1−μ2μ1-μ2 given the following information. n1=45n1=45, ¯x1=2.66x¯1=2.66, s1=0.34s1=0.34 n2=25n2=25, ¯x2=2.89x¯2=2.89, s2=0.71s2=0.71 ±± Use Technology Rounded to 2 decimal places.
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Give a 99.9% confidence interval, for μ1−μ2μ1-μ2 given the following information.
n1=45n1=45, ¯x1=2.66x¯1=2.66, s1=0.34s1=0.34
n2=25n2=25, ¯x2=2.89x¯2=2.89, s2=0.71s2=0.71
±± Use Technology Rounded to 2 decimal places.
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- Give a 99% confidence interval, for μ1−μ2μ1-μ2 given the following information. n1=35n1=35, ¯x1=2.6x¯1=2.6, s1=0.82s1=0.82n2=50n2=50, ¯x2=2.61x¯2=2.61, s2=0.58s2=0.58 ±± Use Technology Rounded to 2 decimal places. HintAs a researcher, you are curious if low income impacts the length of prison sentences for theft. The known average sentence is general population is µ = 12.5 months. You have a random sample of 49 people (N = 49) whose income is below the poverty line. The average sentence for this sample of people was xbar = 16.8 months, with a standard deviation of sx = 4.3. In this case, the critical values for an α (alpha) = .05 are +/- 2.01. A) Please calculate and report the confidence interval for these data, including your level of confidence. Hints: Do not calculate a z/t-statistic. Remember, the interval is centered on a descriptive statistic, such as xbar. The level of confidence is a specific percentage based on alpha; high or low are NOT answers. B) Based on the confidence interval, would you retain or reject the null hypothesis? Explain your reasoning.' C) Would the p-value be bigger or smaller than alpha? D) In everyday language, what do these results tell you about the research…Give a 98% confidence interval, for μ1−μ2 given the following information. n1=45 xbar1=2.62, s1=0.8n2=60n xbar = 3.02, s2=0.67 ____ ±± ____ Use Technology Rounded to 2 decimal places.
- Give a 99.9% confidence interval, for μ1−μ2μ1-μ2 given the following information. n1=50n1=50, ¯x1=2.13x¯1=2.13, s1=0.98s1=0.98n2=40n2=40, ¯x2=1.65x¯2=1.65, s2=0.86s2=0.86 Rounded to 2 decimal places.In the past, a chemical company produced 880 pounds of a certain type of plastic per day. Now, using a newly developed and less expensive process, the mean daily yield of plastic for the first 50 days of production is 871 pounds; the standard deviation is 21 pounds. Do the data provide sufficient evidence to indicate that the mean daily yield for the new process is less than that of the old procedure? (Use α=0.05) (c) p-value for the test above isA random sample of 9 size AA batteries for toys yield a mean of 3.66 hours with standard deviation, 1.05 hours. (a) Find the critical value, t, for a 99% CI. t = (b) Find the margin of error for a 99% CI.
- A random sample of 11 size AA batteries for toys yield a mean of 2.98 hours with standard deviation, 1.19 hours. (a) Find the critical value, t?/2, for a 99% CI. ??/2 = (b) Find the margin of error for a 99% CI.The mean blood cholesterol level for all men aged 20 to 34 years is μ=188μ=188 mg/dl. We suspect that the mean for cross-country runners is lower. We take a random sample from all cross-country runners and measure the cholesterol level of each runner. We set α=0.10α=0.10 and correctly calculate a p−valuep−value of 0.08. Suppose someone else has additional information unavailable to us and it is known that IN REALITY cross-country runners actually do have lower cholesterol levels. Based on our calculations, what conclusion did we reach and did we make an error? The mean blood cholesterol level for all men aged 20 to 34 years is μ=188μ=188mg/dl. We suspect that the mean for cross-country runners is lower. We take a random sample from all cross-country runners and measure the cholesterol level of each runner. We set α=0.10α=0.10and correctly calculate a p−valuep−valueof 0.08. Suppose someone else has additional information unavailable to us and it is known that IN REALITY…A study of working actors looked at age and gender. One sample of 55 male actors had a mean age of 34 and a standard deviation of 3. The other sample included 55 female actors with a mean age of 40 and a standard deviation of 2. Estimate with 95% confidence the difference between the average ages of working male (μ1μ1) and female (μ2μ2) actors. Round answers to the nearest hundredth.< μ1−μ2<
- The additional growth of plants in one week are recorded for 11 plants with a sample standard deviation of 2 inches and sample mean of 12 inches. t* at the 0.05 significance level = Ex: 1.234 Margin of error = Ex: 1.234 Confidence interval = [ Ex: 12.345 E, Ex: 12.345 ]Give a 99% confidence interval, for u1- µ2 given the following information. nį = 30, 1= 2.95, s1 = 0.3 Tg = 35, 2 = 3, s2 = 0.53 土 Use Technology Rounded to 2 decimal places.find a 95% confidence interval for each pair difference. Recall that a confidence interval for μi−μj can be constructed as follows. ( ̄xi− ̄xj)±t∗√MSE(1ni+1nj) TheMSE is the mean square error from the ANOVA table. The t^* is from the t-distribution with n−k degrees of freedom, where n is the total sample size and k the number of groups. Between Low and Medium levels of calcium. Between Low and High levels of calcium. Between Medium and High levels of calcium. Calcium GillRate Low 55 Low 63 Low 78 Low 85 Low 65 Low 98 Low 68 Low 84 Low 44 Low 87 Low 48 Low 86 Low 93 Low 64 Low 83 Low 79 Low 85 Low 65 Low 88 Low 47 Low 68 Low 86 Low 57 Low 53 Low 58 Low 47 Low 62 Low 64 Low 50 Low 45 Medium 38 Medium 42 Medium 63 Medium 46 Medium 55 Medium 63 Medium 36 Medium 58 Medium 73 Medium 69 Medium 55 Medium 68 Medium 63 Medium 73 Medium 45 Medium 79 Medium 41 Medium 83 Medium 60 Medium 48…