Find the critical value t/α2 needed to construct a confidence interval of the given level with the given sample size. Round the answer to at least three decimal places.Level 80%, sample size 18.Critical=value

Find the critical value z/α2 needed to construct a confidence interval with level 80%.Round the answer to two decimal places.The critical value for the 80% confidence level is .

For a confidence level of 80% with a sample size of 22, find the critical t value. Rounding to 3 decimal places is acceptable.

Suppose you want to compute a confidence interval for a population proportion (p).For a confidence level of 92%, find the critical value

For the given confidence level and values of x and n, find the following.=x47, =n99, confidence level 80%Part: 0 / 30 of 3 Parts CompletePart 1 of 3(a) Find the point estimate. Round the answers to at least four decimal places, if necessary.The point estimate for the given data is .

You have taken a random sample of size =n22 from a normal population that has a population mean of =μ95 and a population standard deviation of =σ8. Your sample, which is Sample 1 in the table below, has a mean of =x93.9. (In the table, Sample 1 is written "S1", Sample 2 is written "S2", etc.)(a)Based on Sample 1, graph the 75% and 90% confidence intervals for the population mean. Use 1.150 for the critical value for the 75% confidence interval, and use 1.645 for the critical value for the 90% confidence interval. (If necessary, consult a list of formulas.)Enter the lower and upper limits on the graphs to show each confidence interval. Write your answers with one decimal place.For the points ( and ), enter the population mean, =μ95.75% confidence interval87.0102.0 90% confidence interval87.0102.0(b)Press the "Generate Samples" button below to simulate taking 19 more samples of size =n22 from the population. Notice that the confidence intervals for these samples are drawn automatically. Then complete parts (c) and (d) below the table.x 75%lowerlimit 75%upperlimit 90%lowerlimit 90%upperlimitS1 93.9 ? ? ? ?S2 94.8 92.8 96.8 92.0 97.6S3 94.5 92.5 96.5 91.7 97.3S4 98.7 96.7 100.7 95.9 101.5S5 94.3 92.3 96.3 91.5 97.1S6 94.6 92.6 96.6 91.8 97.4S7 94.6 92.6 96.6 91.8 97.4S8 95.8 93.8 97.8 93.0 98.6S9 97.4 95.4 99.4 94.6 100.2S10 96.1 94.1 98.1 93.3 98.9S11 92.6 90.6 94.6 89.8 95.4S12 96.0 94.0 98.0 93.2 98.8S13 94.5 92.5 96.5 91.7 97.3S14 91.1 89.1 93.1 88.3 93.9S15 96.2 94.2 98.2 93.4 99.0S16 91.4 89.4 93.4 88.6 94.2S17 96.0 94.0 98.0 93.2 98.8S18 94.1 92.1 96.1 91.3 96.9S19 94.2 92.2 96.2 91.4 97.0S20 94.1 92.1 96.1 91.3 96.975% confidence intervals87.0102.090% confidence intervals87.0102.0(c)Notice that for =172085% of the samples, the 90% confidence interval contains the population mean. Choose the correct statement. When constructing 90% confidence intervals for 20 samples of the same size from the population, exactly 90% of the samples will contain the population mean. When constructing 90% confidence intervals for 20 samples of the same size from the population, at most 90% of the samples will contain the population mean. When constructing 90% confidence intervals for 20 samples of the same size from the population, it is possible that more or fewer than 90% of the samples will contain the population mean.(d)Choose ALL that are true. The 90% confidence interval for Sample 8 does not indicate that 90% of the Sample 8 data values are between 93.0 and 98.6. The 75% confidence interval for Sample 8 is narrower than the 90% confidence interval for Sample 8. This must be the case, because when a confidence interval is constructed for a sample, the greater the level of confidence, the wider the confidence interval. From the 75% confidence interval for Sample 8, we know that there is a 75% probability that the population mean is between 93.8 and 97.8. If there were a Sample 21 of size =n44 taken from the same population as Sample 8, then the 90% confidence interval for Sample 21 would be narrower than the 90% confidence interval for Sample 8. None of the choices above are true.