Math  /  Data & Statistics

QuestionA random sample of 40 students has a mean annual earnings of $3120\$ 3120 and assume that the population standard deviation is $677\$ 677. Construct the confidence interval for the population mean, μ\mu if c=0.95\mathrm{c}=0.95. A. ($4812,$5342)(\$ 4812, \$ 5342) B. ($210,$110)(\$ 210, \$ 110) C. ($1987,$2346)(\$ 1987, \$ 2346) D. ($2910,$3330)(\$ 2910, \$ 3330)

Studdy Solution
Construct the confidence interval using the formula:
xˉ±z×SEM \bar{x} \pm z \times \text{SEM}
Substitute the values:
3120±1.96×107.06 3120 \pm 1.96 \times 107.06
3120±209.84 3120 \pm 209.84
Calculate the interval:
Lower limit:
3120209.84=2910.16 3120 - 209.84 = 2910.16
Upper limit:
3120+209.84=3329.84 3120 + 209.84 = 3329.84
Round to the nearest dollar:
(2910,3330) (2910, 3330)
The confidence interval for the population mean is:
($2910,$3330) \boxed{(\$ 2910, \$ 3330)}

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