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A GROUP OF STATISTICS STUDENTS DECIDED TO CONDUCT A SURVEY AT THEIR UNIVERSITY TO FIND THE AVERAGE (MEAN) AMOUNT OF TIME STUDENTS SPENT STUDYING PER WEEK. THEY SAMPLED 240 STUDENTS AND FOUND A MEAN OF 22.3 HOURS PER WEEK. ASSUMING A POPULATION STANDARD DEVIATION OF SIX HOURS, WHAT IS 99% LEVEL OF CONFIDENCE?A. [21.80, 22.80]B. [16.3, 28.3]C. [21.30, 23.30]D. [20.22, 22.0]

Respuesta :

Answer: Option 'c' is correct.

Step-by-step explanation:

Since we have given that

n = 240

mean = 22.3 hours

standard deviation = 6 hours

At 99% level of confidence, z = 2.58

So, interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=22.3\pm 2.58\times \dfrac{6}{\sqrt{240}}\\\\=22.3\pm 0.99\\\\=(22.3-0.99,22.3+0.99)\\\\=(21.31,23.29)[/tex]

Hence, Option 'c' is correct.

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