Question

Last year the commercial time for a broadcast on a particular TV channel was normally distributed with
Expectation 6.1 minutes. The channel manager claims that this year there has been a change in the life expectancy of the broadcast time.
To test his claim, he modeled 16 hours of transmission and found that the average commercial time per hour of transmission
Is 5.7 hours with a standard deviation of 1.2 hours.

A. Would you justify the channel manager's claim with a significance level of 0.05 ?

B. What is the smallest level of significance by which to justify his claim?

C. The channel's VP of marketing built the following trust profit according to the same sample:
p {4.92 <μ <6.48} = 1-α, what level of trust did you use?

D. The elapsed time between two commercials is split exponentially with a span of 25 minutes.
What is the probability that more than 20 minutes will pass between 2 advertisements if less than 30 are known to have passed
subtlety?

Last year the commercial time for a broadcast on a particular TV channel was normally distributed with
Expectation 6.1 minutes. The channel manager claims that this year there has been a change in the life expectancy of the broadcast time.
To test his claim, he modeled 16 hours of transmission and found that the average commercial time per hour of transmission
Is 5.7 minutes with a standard deviation of 1.2 minutes.

A. Would you justify the channel manager's claim with a significance level of 0.05 ?

B. What is the smallest level of significance by which to justify his claim?

C. The channel's VP of marketing built the following trust profit according to the same sample:
p {4.92 <μ <6.48} = 1-α, what level of trust did you use?

D. The elapsed time between two commercials is split exponentially with a span of 25 minutes.
What is the probability that more than 20 minutes will pass between 2 advertisements if less than 30 are known to have passed
subtlety?

Answer 1

Expected commercial time for adverisement = = 6.1 mins

Here sample size = n= 16

sample mean = = 5.7 hrs

sample standar deviation = s = 1.2 hours

standard error = s/sqrt(n) = 1.2/sqrt(16) = 0.3 hours

H₀: = 6.1 mins

H_a : < 6.1 mins

Test statistic

t = (5.7 - 6.1)/0.3 = -1.333

Here critical value = TINV(0.05, dF = 16-1 = 15) = 2.13145

so here l t l < t_critical so we would fail to reject the null hypothesis.

(b) Here

p -value = 2 * TDIST(t < -1.3333; dF = 15) = 0.2023

so here smallest level of significance by which to justify his claim is 0.2023.

(c) p {4.92 <μ <6.48} = 1-α

t₁ = (4.92 - 6.1)/0.3 = -3.933

t₂ = (6.48 - 6.1)/0.3 = 1.2667

p {4.92 <μ <6.48} = TDIST(t < 1.2667; dF = 15) - TDIST(t < -3.9333; dF = 15)

= 0.1123 - 0.0007 = 0.1116

α =1 - 0.1116 = 0.8884

(d) Here in the past whateve happen it doesn't impact on the future in exponential distribution.

f(t) = (1/25) e^-t/25 ; t > 0

P(t > 20) = 1-(1 - e^-20/25)  = 0.4493