2:30 AM famu.blackboard.com Verizon LTE 46% . 1) Find the critical z-value_ a 72

Question

2:30 AM famu.blackboard.com Verizon LTE 46% . 1) Find the critical z-value_ a 72 % confidence interval for , the population mean, with known. Ans 1.08 2) Find the critical z-value-a 92 % confidence interval for , the population mean, with known. Ans= 1.75 3) Construct a 90% confidence interval for . The known population standard deviation =10. The sample data is{-=60and n = 4) Construct a 95% confidence interval for . The known population standard deviation =20. The sample data is -90and n- 300 5). A confidence interval for has the form- E where E is the error term. Find the minimum sample size n that must be taken so that E 2 for an 80% confidence interval where the population standard deviation 16. 6) A normally distributed population has an unknown standard deviation. The sample data is {=20' s = 6 and n-5} Find the t-value to construct a 95% confidence interval for the population mean . 7) A normally distributed population has an unknown standard deviation. The sample data is { -16, s 8 and n = 67) Find the t-value to construct a 90% confidence interval for the population mean . 8) A normally distributed population has an unknown standard deviation. The sample data is { = 30, s = 8 and n-20} Construct an 80% confidence interval for the population mean . 9) A normally distributed population has an unknown standard deviation. The sample data is { = 40, s = 10 and n = 160) Construct a 90% confidence interval for the population mean .

Explanation / Answer

1.       Critical value of 72% confidential interval (CI) for population mean (µ) can be estimated as follows:

Estimate = 1/100*(100-CI) and then /2 (since the problems referred here are two-tailed tests)

The area left to this in normal distribution is 1- /2. Then in z-table, look for the z-score, which is the critical value.

Here, CI= 72%, /2 = 0.14; 1- /2 = 0.86. The z-value from table = 1.08.

You can also use the excel-command NORMSINV(1- /2) and get the same result

2.       Follow the same procedure for CI = 92%. Then 1- /2 = 0.04. The z-value from table = 1.75.

You can also use the excel-command NORMSINV(1- /2) and get the same result.

3.       Given : Sample mean (x-bar) = 60, No. of samples n=80, Standard deviation()=10. Here, /2 (for 90% confidence) = 0.05, z-score for /2 = 0.05 is 1.645 (see the table)

Estimated confidence interval = mean ± Z /2(/ n)= 60±1.839 (i.e., 60± 1.84)

4.       Following the same procedure, for X-bar =90, n=300, =20 & /2 = 0.025 (=>Z /2=1.96)

Estimated confidence interval = mean ± Z /2(/ n)= 90±2.263 (i.e., 90± 2.26)

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