Question

The following data represent a random sample of the ages of players in a baseball league. Assume that the population is normally distributed with a standard deviation of 1.8 years. Find the 95% confidence interval for the true mean age of players in this league. Round your answers to two decimal places and use ascending order.

Age: 32, 24, 30,34,28, 23,31,33,27,25

Answer 1

(27.579, 29.104)

Explanation:

A 98% confidence interval for μ is (x¯−zα/2σn‾√,x¯+zα/2σn‾√). Here, α=0.02, σ=2.1, and n=41. Use Excel to calculate the 98% confidence interval.1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean, rounded to five decimal places, is x¯=28.34146.2. Click on any empty cell, enter =CONFIDENCE.NORM(0.02,2.1,41), and press ENTER.3. The margin of error, rounded to five decimal places is zα/2σn‾√≈0.76296. The confidence interval for the population mean has a lower limit of 28.34146−0.76296=27.57850 and an upper limit of 28.34146+0.76296=29.10442.Thus, the 98% confidence interval for μ is (27.579, 29.104).

Answer 2

Explanation:

n = 10

Mean, m = (32 + 24 + 30 + 34 + 28 + 23 + 31 + 33 + 27 + 25)/10

= 28.7

From the information given,

Standard deviation, s = 1.8

For a confidence level of 95%, the corresponding z value is 1.96.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

28.7 ± 1.96 × 1.8/√10

= 28.7 ± 1.12

The lower end of the confidence interval is 28.7 - 1.12 = 27.58 years

The upper end of the confidence interval is 28.7 + 1.12 = 29.82 years