Mathematics: Suppose that the scores of a standardized test are normally distributed with an unknown mean and standard deviation. A random sample of 21 scores is taken and gives a sample mean of 95 points and a sample standard deviation of 8 points. df t0.10 t0.05 t0.025 t0.01 t0.005 17 1.333 1.740 2.110 2.567 2.898 18 1.330 1.734 2.101 2.552 2.878 19 1.328 1.729 2.093 2.539 2.861 20 1.325 1.725 2.086 2.528 2.845 21 1.323 1.721 2.080 2.518 2.831 Find the margin of error for a 95% confidence interval estimate

Let the age of Kevin be x,age of twin sisters = x - 6sum of children s age = x + x - 6 + x - 6= 3x - 123x - 12 = 423x = 54x = 54/3= 18age of twin sisters = x - 6= 18 - 6= 12Kevin is 18 years old whereas the twin sisters are both 12 years old....

Suppose that the scores of a standardized test

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notsosmart249

29.01.2022

Mathematics

ME = 3.64

Step-by-step explanation:

Given:

Sample size, n = 21

Sample mean, X' = 95

Standard deviation $\sigma$ = 8.

Confidence interval = 95%

Significance level $\alpha$ = 0.05

Required:

Find margin of error

To find the margin of error, use the formula below:

$M.E = \frac{\sigma}{\sqrt{n}} * t_\alpha_/_2; _d_f$

Where,

$t_\alpha_/_2 = 0.05/2 = 0.025$

degree of freedom, df = n - 1 = 21 - 1 = 20

Therefore,

$M.E = \frac{8}{\sqrt{21}} * t_0_._0_2_5; _2_0$

Using the table at 95% Confidence interval, we have:

$= \frac{8}{\sqrt{21}} * 2.086$

= 3.6416

Rounding to 2 decimal places,

ME = 3.64

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