The amount of time adults spend watching television is closely monitored by firms becayse this helps to determine advertising pricing for commercials. Compete parts (a) through (d).
a) Do you think the variable "weekly time spent watching television" would be normally distributed? Yes or No.
If not, what shape would you expect the variable to have? Skewed Left, Skewed Right, Uniform or Symmetric?
b) According to a certain survey, adults spend 2.35 hours per day watching television on a weekday. Assume that the standard deviation for "time spent watching TV on a weekday" is 1.93 hours. If a random sample of 40 adults is obtained, describe the sampling distribution of x-bar, the mean amount of time spent watching TV on a weekday.
Mean =
A) 2.35
B) 1.89
C) 2.25
SD = (round to six decimal places as needed)
c) Determine the probability that a random sample of 40 adults results in a mean time watching television on a weekday of between 2 and 3 hours.
The probability is .
d) One consequence of the popularity of the Internet is that it is thought to reduce TV watching. Suppose that a random sample of 35 people who consider themselves avid Internet users results in a mean time of 1.89 hours watching TV on a weekday. Determine the likelihood of obtaining a sample mean of 1.89 hours or less from a population who mean is presumed to be 2.35 hours.
The likelihood is .
Based on the result obtained, do Internet users watch less TV? Yes or No.

(0.8559 we check the value of 2.13 from the normal distribution tables and add with the value of 1.14 to get the in between value -1.14694 < Z <2.13003)

d) Here n = 35 , s= 1.93 , mean = 2.35 and x= 1.89

So Putting the values

P (X ≤ 1.89) = P (Z ≤ 1.89- 2.35/ 1.93 / √35)

= P ( Z ≤ -0.238341/ 5.9160797)

= P ( Z ≤ -0.04028)

= 0.5 - 0.0159

= 0.4841

Similarly again subtracting from 0.5 the value from normal distribution table to get less than or equal to value.

Yes based on the result 0.4841 the internet users watch less TV because the mean would be placed at 0.5 and it is less than 0.5

maheshpatelvVT maheshpatelvVT · Answer 2 · 2022-04-07T19:27:10+02:00

The mean value is 2.35, SD is 1.93 with an SD error of 0.3051597 and the probability is 0.8559, and yes 0.4841 internet users watch less TV.

It is given that the amount of time adults spend watching television is closely monitored by firms because this helps to determine advertising pricing for commercials.

It is required to find the standard deviation and probability.

What is a confidence interval for population standard deviation?

It is defined as the sampling distribution following an approximately normal distribution for known standard deviation.

We know the formula for standard error:

[tex]\rm SE = \frac{s}{\sqrt{n} }[/tex]

Where 's' is the standard error

and n is the sample size.

In the question the value of s = 1.93 hours

and sample size n = 40

[tex]\rm SE = \frac{1.93}{\sqrt{40} }\\[/tex]

SE = 0.3051597

For the probability between 2 and 3 hours.

= P(2<X<3)

[tex]\\\rm =P(\frac{2-x)}{s} < Z < \frac{3-x)}{s})\\\\\rm = P(\frac{(2-2.35)}{0.3051598} < Z < \frac{(3-x)}{0.3051598})\\\\[/tex] (because the mean value x is 2.35)

=P(-1.14694 <Z < 2.13003)

=0.3729+ 0.4830 ( values get from Z table for -1.14 and 2.13 )

=0.8559

For P(X≤1.89)

[tex]\rm P(Z\leq \frac{(x'-x)}{\frac{s}{\sqrt[]{n} } } )\\\\\rm P(Z\leq \frac{(1.89-2.35)}{\frac{1.89}{\sqrt[]{35} } } )[/tex]

= P(Z ≤ -0.04028)

= 0.5 - 0.0159

=0.4841

Based on the result of 0.4841 the internet users watch less TV because the mean would be placed at 0.5 and it is less than 0.5.

Thus, the mean value is 2.35, SD is 1.93 with an SD error of 0.3051597 and the probability is 0.8559, and yes internet users watch less TV.

Learn more about the standard deviation here:

https://brainly.com/question/12402189