Find critical value tα/2 for n = 15 and 98% confidence level. A.±2.6245 B.±2.4142 C.±2.4460 D.±2.5551

Find critical value t_α/2 for n = 15 and 98% confidence level.

A.±2.6245

B.±2.4142

C.±2.4460

D.±2.5551

Expert Solution

critical value tα/2 for sample size(n) = 15 and 98% confidence level is

level of significance = 1 - 0.98 = 0.02

degree of freedom = n -1 = 15 - 1 = 14

Hence in t table we will look for value with level of significance = 0.02 (Two tails) and degree of freedom = 14

which is equal to 2.6245.

option A is correct.

please like)

orchestra answered 1 year ago

For a confidence level of 98%, find the critical value for a normally distributed variable. The...

For a confidence level of 98%, find the critical value for a normally distributed variable. The sample mean is normally distributed if the population standard deviation is known.

1a. Find the critical value tc for the confidence level c=0.80 and sample size n=15. tc=...

1a. Find the critical value tc for the confidence level c=0.80 and sample size n=15. tc= 1b. Find the critical value tc for the confidence level c=0.80 and sample size n=21. tc= 1c. Find the margin of error for the given values of c, s, and n. c=0.98,s=4, n=11

1) For a confidence level of 92%, find the Z-critical value 2) If n = 500...

1) For a confidence level of 92%, find the Z-critical value 2) If n = 500 and ˆpp^ (p-hat) = 0.7, construct a 99% confidence interval. 3) Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 405 drivers and find that 299 claim to always buckle up. Construct a 80% confidence interval for the population proportion that claim to always buckle up. 4) A political candidate...

For a confidence level of 98% with a sample size of 28, find the critical t...

For a confidence level of 98% with a sample size of 28, find the critical t value. Critical Value = (round answer to 3 decimal places) Assume that a sample is used to estimate a population mean μμ. Find the 99.9% confidence interval for a sample of size 33 with a mean of 72.3 and a standard deviation of 12.1. Enter your answer accurate to one decimal place. I am 99.9% confident that the mean μμ is between and Assume that...

A) Find the z critical value for a 90% confidence level: Zc= B) Find the z...

A) Find the z critical value for a 90% confidence level: Zc= B) Find the z critical value for an 84% confidence level: Zc= C) The higher the confidence, the narrower the confidence interval True or False D) The greater the sample size, the mnarrower the confidence interval True or False E) In a recent study of 84 eighth graders, the mean number of hours per week that they watched TV was 22.3. Assume population standard deviation is 5.8 hours....

find the critical value Z α/2 when the confidence level is 92%.

1. Find the critical value ?) ??/z 2 , for a 96% confidence level. ?) ??/z...

1. Find the critical value ?) ??/z 2 , for a 96% confidence level. ?) ??/z 2 , for a 99.5% confidence level. ?) ??/z 2 , for a 98% confidence level and for a sample of size 24. ?) ta/z 2 , for a 90% confidence level and for a sample of size 39.

Find the critical value z Subscript c necessary to form a confidence interval at the level...

Find the critical value z Subscript c necessary to form a confidence interval at the level of confidence shown below. zc= 0.87

Find the critical value, x2/R and x2/L for c=.98 and n=20 Show work.

Find the critical z value for a confidence level of 94%. Critical Value= round answer to...

Find the critical z value for a confidence level of 94%. Critical Value= round answer to 2 decimal places) Out of 300 people sampled, 246 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places I am 99% confident that the proportion of people who have kids is between and Out of 300 people sampled, 255 had kids. Based on this, construct a 95%...

Question