1) You are interested in how stressed out students are during finals week. You k

Question

1) You are interested in how stressed out students are during finals week. You know (magically) that the standard deviation ‘sigma’ of all students stress levels is equal to 4 but are interested in learning about the unknown mean mu of all students. So you go out and measure stress levels in 15 students and find that your sample mean X-bar =10.

(a) What distribution is this sample mean drawn from? (2---pts)

(b) What is critical Z* for 95% confidence? (1---pts)

(c) What is the 95% confidence interval? (2---pts)

2) Suppose you did not magically know sigma a) Based on the data you collected what would you use in its place? (1---pts)

b) In doing so, you would no longer be able to use the Z---distribution and instead would have to use the Distribution (2---pt)

c) What would the new critical value be for 95% confidence? (2---pt)

Explanation / Answer

(1) (a) Normal distribution

(b) Z critical for 95% confidence is z = 1.96

(c)

n = 15     

x-bar = 10     

s = 4     

% = 95     

Standard Error, SE = /n =    4 /15 = 1.032795559

z- score = 1.959963985     

Width of the confidence interval = z * SE =     1.95996398454005 * 1.03279555898864 = 2.024242099

Lower Limit of the confidence interval = x-bar - width =      10 - 2.02424209901066 = 7.975757901

Upper Limit of the confidence interval = x-bar + width =      10 + 2.02424209901066 = 12.0242421

The 95% confidence interval is [7.98, 12.02]

(2) (a) We would use the sample standard deviation

(b) t- distribution

(c) t critical for 95% confidence and 14 degrees of freedom is t = 2.145

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