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One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 45 individuals who consider themselves to be avid Internet users results in a mean time of 1.98 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.98 hours or less from a population whose mean is presumed to be 2.35 hours.

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Nuhulawal20idn

This question is incomplete, the complete question is;

One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 45 individuals who consider themselves to be avid Internet users results in a mean time of 1.98 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.98 hours or less from a population whose mean is presumed to be 2.35 hours.

Assume that the standard deviation for​ "time spent watching television on a​ weekday" is 1.93 hours.

Answer:

the likelihood of obtaining a sample mean of 1.98 hours or less is  0.0985  

Step-by-step explanation:

Given the data in the question;

x" = 1.98  

μ = 2.35  

s = 1.93

n = 45

so, p( x" < 1.98 ) = P( (x" - μ)/(s/√n) )

we substitute in our values,

p( x" < 1.98 ) = P( (1.98 - 2.35) / (1.93/√45) )

= p( -0.37 / 0.2877

= P(  - 1.29 )

from standard normal table,

P(  - 1.29 ) =  0.0985  

Therefore, the likelihood of obtaining a sample mean of 1.98 hours or less is  0.0985  

 

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