Determine the sample size required to estimate the mean score on a standardized test within 4 points of the true mean with 98% confidence. Assume that s = 14 based on earlier studies.

Formula to calculate sample size: [tex]n=(\dfrac{z^*\times s}{E})^2[/tex] , where s = standard deviation on prior studies , z* = two tailed critical value for confidence level , E = margin of error

Given: E = 4 , s= 14 and confidence level = 98%

Z-value for 98% confidence = 2.326

Then, required sample size: [tex]n=(\dfrac{2.326\times 14}{4})^2=(8.141)^2=66.275881\approx67[/tex]

Hence, the required sample size = 67 .

raphealnwobi raphealnwobi · Answer 2 · 2021-11-23T11:23:49+01:00

Sample size required to estimate the mean score on a standardized test is 67.

The confidence level (C) = 98% = 0.98

α = 1 - C = 0.02

α/2 = 0.02/2 = 0.01

The z score of α/2 is the same as the z score of 0.49 (0.50 - 0.01) which is equal to 2.33

Given that:

Standard deviation (σ) = 14, margin of error (E) = 4, n = sample size, hence:

[tex]E=z_\frac{\alpha}{2}*\frac{\sigma}{\sqrt{n} } \\\\4=2.33*\frac{14}{\sqrt{n} } \\\\\sqrt{n}=8.155\\\\n=67[/tex]

Hence the sample size required to estimate the mean score on a standardized test is 67.

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