The same radio station wants to track the average number of commercial minutes i

Question

The same radio station wants to track the average number of commercial minutes it plays in an hour. A random sample over the past month yielded the following data in minutes: 15.5, 10.2, 8.2, 16, 14.4, 9.5, 8.9, 11.4, 16.3, and 15.2.

A. Compute a 99% confidence interval for the mean number of commercial minutes played in an hour. (2 points)

B. The station manager is concerned that listeners will change the station if too many commercials are played per hour. The program manager says the show plays less than 15 minutes of commercials per hour. Set up the hypotheses to test the program manager's statement. (1 points)

C. What test will you use to evaluate the alternative hypothesis? What are the conditions of this test and are they met in this situation? (1 points)

D. Calculate your test statistic and P-value. Show your work. (2 points)

E. What's your critical t value and conclusion at alpha = .05? (2 points)

Explanation / Answer

a.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=12.56
Standard deviation( sd )=3.2232
Sample Size(n)=10
Confidence Interval = [ 12.56 ± t a/2 ( 3.2232/ Sqrt ( 10) ) ]
= [ 12.56 - 3.25 * (1.019) , 12.56 + 3.25 * (1.019) ]
= [ 9.247,15.873 ]

b & c.
Set Up Hypothesis
Null, H0: U=15
Alternate,show plays less than 15 minutes of commercials per hour H1: U<15

d & e.
Test Statistic
Population Mean(U)=15
Sample X(Mean)=12.56
Standard Deviation(S.D)=3.2232
Number (n)=10
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =12.56-15/(3.2232/Sqrt(10))
to =-2.394
| to | =2.394
Critical Value
The Value of |t | with n-1 = 9 d.f is 1.833
We got |to| =2.394 & | t | =1.833
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Left Tail -Ha : ( P < -2.3939 ) = 0.02015
Hence Value of P0.05 > 0.02015,Here we Reject Ho

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