Question 1182978: Assume that we want to construct a confidence interval. (a) Find the critical value t α/2, (b) find the critical value z α/2 or (c) state that neither the normal distribution nor the t distribution applies.
Here are summary statistics for randomly selected weights of newborn girls: n = 169, x = 29.2 hg, s = 7.3 hg. The confidence level is 95%
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! I would use a t-test since the sd of the sample is used as a variability estimator. It is true that n is > 30, so the difference between z and t is < 1%. Furthermore, given the ease of calculators today, dealing with a t-table is essentially the same as dealing with a z-table.
The critical z-value (0.975) is 1.96; the t-value (0.975, df=168) is 1.97. My answer would be a, but some would say use z. The difference is small.
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the half-interval for the interval is t*s/sqrt(n)
=1.97*7.3/13
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