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svp [43]
3 years ago
14

What will the ladder reach?

Mathematics
1 answer:
EastWind [94]3 years ago
7 0
I would say approximately 8ft. <span />
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A. Use the​ one-mean t-interval procedure with the sample​ mean, sample​ size, sample standard​ deviation, and confidence level
PtichkaEL [24]

Answer:

a) (17.227, 22.773)

b) 2.773

c) 2.773

Step-by-step explanation:

Given:

Sample size, n = 25

Standard deviation, s = 4

Sample mean, x' = 20

Level of significance, a = 0.98 = 1 - 0.98 = 0.02

The degrees of freedom, df, for a t-distribution = n - 1 = 25 - 1 = 24

Using the t table, the Critical value = t_\alpha _/_2, _d_f = t_0_._0_2_/_2, _2_4 = t_0_._0_1_, _2_4 = 3.4668

Margin of error, E = t_\alpha _/_2, _d_f * \frac{\sigma}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

Limits of 98% confidence interval, we have:

Lower limit : x' - M.E = 20 - 2.773 = 17.227

Upper limit: x' + M.E = 20 + 2.773 = 22.773

Therefore, (17.227, 22.773) is 98% confidence interval.

b) Let's the margin of error by taking half the length of the confidence interval.

Since we are to use half the length of CI, we have:

M.E = \frac{22.773 - 17.227}{2} = 2.773

c)M.E = t_\alpha _/_2, _d_f *s/\sqrt{n}

= t * \frac{s}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

4 0
3 years ago
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