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 for the population mean using the Student's t-distribution
Use the portion of the table above. Round the final answer to two decimal places.
Provide your answer below:
ME = _______.

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Chrisnando Chrisnando · Answer 1 · 2020-07-09T03:35:51+02:00

Answer:

ME = 3.64

Step-by-step explanation:

Given:

Sample size, n = 21

Sample mean, X' = 95

Standard deviation [tex] \sigma [/tex] = 8.

Confidence interval = 95%

Significance level [tex] \alpha[/tex] = 0.05

Required:

Find margin of error

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

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

Where,

[tex] t_\alpha_/_2 = 0.05/2 = 0.025[/tex]

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

Therefore,

[tex] M.E = \frac{8}{\sqrt{21}} * t_0_._0_2_5; _2_0 [/tex]

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

[tex] = \frac{8}{\sqrt{21}} * 2.086 [/tex]

[tex] = 3.6416 [/tex]

Rounding to 2 decimal places,

ME = 3.64