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3 years ago

Pls help i really need to get this done

Mathematics

2 answers:

prisoha [69]3 years ago

8 0

The answer is “C”. Have a good day!

nataly862011 [7]3 years ago

6 0

The answer to this is c

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3 years ago

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4 years ago

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

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

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3 years ago

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