Scores on a recent Statistics test are normally distributed with a standard deviation of 6.5 points. If the professor wants to estimate the population mean test score within 1 point with 90% confidence, what sample size is needed

A sample of size =n90 is drawn from a normal population whose standard deviation is =σ9.7. The sample mean is =x38.78.Part: 0 / 20 of 2 Parts CompletePart 1 of 2(a) Construct an 80% confidence interval for μ. Round the answer to at least two decimal places.An 80% confidence interval for the mean is <<μ.

You want to compute a 90% confidence interval for the mean of a population with unknown standard deviation. The sample size is 30. The correct distribution value for calculating the required confidence interval is:-Group of answer choices1.961.64491.6991.6971.311

A population's standard deviation is 12. We want to estimate the population mean with a margin of error of 2, with a 95% level of confidence. (Use t Distribution Table & z Distribution Table.)How large a sample is required? (Round your intermediate calculations to 2 decimal places and round up your answer to the next whole number.)

A normal population has mean 100 and variance 25.How large must the random sample be if you want the standard error of the sample average to be 1.5?a.9b.10c.11d.None of the otherse.12

Suppose we are interested in studying a population to estimate its mean. The population is normal and has a standard deviation of =σ23. We have taken a random sample of size =n95 from the population. This is Sample 1 in the table below. (In the table, Sample 1 is written "S1", Sample 2 is written "S2", etc.) As shown in the table, the sample mean of Sample 1 is =x140.9. Also shown are the lower and upper limits of the 80% confidence interval for the population mean using this sample, as well as the lower and upper limits of the 95% confidence interval. Suppose that the true mean of the population is =μ140, which is shown on the displays for the confidence intervals.Press the "Generate Samples" button to simulate taking 19 more random samples of size =n95 from this same population. (The 80% and 95% confidence intervals for all of the samples are shown in the table and graphed.) Then complete parts (a) through (c) below the table.x 80%lowerlimit 80%upperlimit 95%lowerlimit 95%upperlimitS1 140.9 138.2 143.6 136.3 145.5S2 139.9 137.2 142.6 135.3 144.5S3 141.2 138.5 143.9 136.6 145.8S4 139.7 137.0 142.4 135.1 144.3S5 143.6 140.9 146.3 139.0 148.2S6 141.5 138.8 144.2 136.9 146.1S7 139.0 136.3 141.7 134.4 143.6S8 136.5 133.8 139.2 131.9 141.1S9 141.8 139.1 144.5 137.2 146.4S10 141.2 138.5 143.9 136.6 145.8S11 145.2 142.5 147.9 140.6 149.8S12 141.5 138.8 144.2 136.9 146.1S13 136.4 133.7 139.1 131.8 141.0S14 142.0 139.3 144.7 137.4 146.6S15 140.2 137.5 142.9 135.6 144.8S16 143.6 140.9 146.3 139.0 148.2S17 138.5 135.8 141.2 133.9 143.1S18 134.7 132.0 137.4 130.1 139.3S19 139.6 136.9 142.3 135.0 144.2S20 141.7 139.0 144.4 137.1 146.380% confidence intervals130.0150.095% confidence intervals130.0150.0(a)How many of the 80% confidence intervals constructed from the 20 samples contain the population mean, =μ140? (b)How many of the 95% confidence intervals constructed from the 20 samples contain the population mean, =μ140? (c)Choose ALL that are true. The center of the 80% confidence interval for Sample 1 is 140.9, because the center of a confidence interval for the population mean must be the sample mean. For each sample, the 80% confidence interval for the sample is included in the 95% confidence interval for the sample. Since Sample 19 and Sample 20 are drawn from the same population, the center of the 95% confidence interval for Sample 19 must be the same as the center of the 95% confidence interval for Sample 20. All of the 95% confidence intervals should be the same as each other. Since they are not all the same, there must have been errors due to rounding. None of the choices above are true.