Imagine we sampled 9 people. The sample mean of cigarettes per week was 5. The sample...

Imagine we sampled 9 people. The sample mean of cigarettes per week was 5. The sample standard deviation was 2. What is the 99% confidence interval for the true mean of cigarettes per week? Show work.

Solutions

Expert Solution

Solution :

Given that,

= 5

s = 2

n = 9

Degrees of freedom = df = n - 1 = 9 - 1 = 8

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t_/2 df} = t_0.005,8 = 3.355

Margin of error = E = t_/2,df * (s / $\sqrt$ n)

= 3.355 * (2 / $\sqrt$ 9) = 2.24 (rounded)

The 99% confidence interval is,

- E < < + E

5-2.24 < < 5+2.24

2.76< < 7.24

99% confidence interval for the true mean of cigarettes per week ( 92.76,7.24)

orchestra answered 1 year ago

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