Question

- From a preliminary survey of 35 students as to how many times they go to a gym during the week, the standard deviation was found to be 3.6. If you want to construct a 95% confidence interval for the mean number of times per week students visit a gym and have a margin of error within .05, what is the minimum number of additional students that should be surveyed?

- What would be the degrees of freedom for a sample size of 48?

Answer #1

Solution:

Given ,

= 3.6 ..Population SD

E = 0.05 Margin of error

c = 95% = 0.95 ...confidence level

Find sample size required.

c = 0.95

= 1- c = 1- 0.95 = 0.05

/2 = 0.025

Using Z table ,

= 1.96

Now, sample size (n) is given by,

= {(1.96* 3.6)/ 0.05}^{2}

= 19914.8544

=19915 ..(round to the next whole number)

New sample size = 19915

Old sample size = 35

19915 - 35 = 19880

**Additional 19880 students that should be
surveyed.**

What would be the degrees of freedom for a sample size of 48?

degrees of freedom = n - 1 = 48 - 1 = 47

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