You want to estimate the mean weight loss of people one year after using a popular weight-loss program being advertised on TV. How many people on that program must be surveyed if we want to be 95% confident that the sample mean weight loss is within 0.25 lb of the true population mean? Assume that the population standard deviation is known to be 10.6 lb.

A. 6907
B. 4865
C. 84
D. 6906

Respuesta :

The number of people on that program is 6907 A

Step-by-step explanation:

The formula of the sample size n is [tex]n=[\frac{p.s}{E}]^{2}[/tex] , where

  • P is the critical value of the confidence interval
  • s is the population standard deviation
  • E is the margin of error

∵ You want to be 95% confident

∵ The critical value of 95% is 1.96

∴ p = 1.96

- We get it by these steps:

  1. 1 - 95% = 1 - 0.95 = 0.05
  2. Divide 0.05 by 2 ⇒ 0.05 ÷ 2 = 0.025
  3. Search for this value in the t table to find p which equivalent to 0.025, you will find it 1.960 (t table is attached)

∵ The sample mean weight loss is within 0.25 lb of the true

   population mean

- That means the margin of error is 0.25

∴ E = 0.25

∵ The population standard deviation is 10.6 lb

∴ s = 10.6

- Substitute these values in the formula of n above

∵ [tex]n=[\frac{(1.96).(10.6)}{0.25}]^{2}[/tex]

∴ n = 6906.2748

- Remember we round the number up

∴ n ≅ 6907

The number of people on that program is 6907

Learn more:

You can learn more about the sample size in brainly.com/question/2826639

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