Question

9. In a certain management study, 15 randomly selected managers were

   found to spend a mean of 2.40 hours each day on paperwork with a standard deviation of 1.30 hours.

   1. Construct a 90% Confidence Interval for the mean time spent on paperwork by all managers.

   2. State the meaning of the interval in the context of the problem.

10. Use the given data to construct a confidence interval at the requested level:

    x = 125, n = 317, confidence level 95% (Hint: Refer to # Va and VI from Test 3.)

11. AtestismadeofH0: μ=14versusHa: μ≠14. Asampleofsize
    n = 48 is drawn, and the mean of the sample equals 12. The population standard deviation is σ = 6.
    a. Is H0 rejected at the α = 5% level? ______________Explain your

    answer by referring to the six-step hypothesis test method

12. a. Find the critical values for a 90% confidence interval using the chi-square distribution with df = 15

    b. Now construct the 90% confidence interval with s2 = 55.5 and n = 16, df = 15.

IX. In a certain management study, 15 randomly selected managers were found to spend a mean of 2.40 hours each day on paperwork with a standard deviation of 1.30 hours. a. Construct a 90% Confidence Interval for the mean time spent on paperwork by all managers. b. State the meaning of the interval in the context of the problem. X. Use the given data to construct a confidence interval at the requested level: x= 125, n = 317, confidence level 95% (Hint: Refer to #Va and VI from Test 3.) XI. A test is made of Ho : u = 14 versus Hą: u # 14. A sample of size n= 48 is drawn, and the mean of the sample equals 12. The population standard deviation is o = 6. a. Is Ho rejected at the a = 5% level? Explain your answer by referring to the six-step hypothesis test method XII. a. Find the critical values for a 90% confidence interval using the chi-square distribution with df = 15 b. Now construct the 90% confidence interval with s2 = 55.5 and n = 16, df = 15.

Answer 1

Given :

Sample size = n = 15

Sample mean = ${\bar{x}}$ = 2.40

Standard deviation = s = 1.30

Significance level = $\alpha$ = 1-0.90 = 0.10

Degree of freedom = df = n-1 = 15-1 = 14

At 90% confidence level, the critical value of t is

taiz, df = sto:1012,143 1.761 .: 90o). CI is CI-X + td 12,df XS xo = 2.40 +1.761 X 1.30 15 = 2.40+ 0.5911 - (1.8089, 2.9911)

We are 90% confident that 90% Confidence Interval for the mean time spent on paperwork by all managers is between 1.8089 and 2.9911

2 ouwen - n = 317 ,x=125 125 Sample proportion: P = = 0.3943 317 critical value: Zx12 - 20.0512=1.96 :95% CI: CF = P + Z0128 PCI-B) co-3943 +1.9610.3943(1-0.3943) 317 -0.394 3 10.0538 - (0.3405, 0.4481)

Which is the required 95% confidence interval.

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