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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

STEP 1

1. The sample size is n=40 n = 40 .
2. The sample mean is xˉ=3120 \bar{x} = 3120 .
3. The population standard deviation is σ=677 \sigma = 677 .
4. The confidence level is c=0.95 c = 0.95 .

STEP 2

1. Determine the z-score for the given confidence level.
2. Calculate the standard error of the mean.
3. Construct the confidence interval for the population mean.

STEP 3

Find the z-score for a 95% confidence level. For a 95% confidence interval, the z-score is approximately 1.96 1.96 .

STEP 4

Calculate the standard error of the mean (SEM) using the formula:
SEM=σn \text{SEM} = \frac{\sigma}{\sqrt{n}}
Substitute the given values:
SEM=67740 \text{SEM} = \frac{677}{\sqrt{40}}
SEM≈6776.3246 \text{SEM} \approx \frac{677}{6.3246}
SEM≈107.06 \text{SEM} \approx 107.06

STEP 5

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:
3120−209.84=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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