Question 847524: 8. A company is experimenting with marketing advertisements delivered by phone calls. Of primary concern is how much of a pre-recorded 7-minute advertisement is heard. In a random sample of 16 calls a mean of 1.85 minutes of the ad was ‘heard’ with a standard deviation of 1.4 minutes. An employee, who had taken STAT 200 while in college, suggested the mean length of time the advertisement is heard would increase if a catchy tune is played softly throughout the advertisement. To test this employee’s claim (although he/she should be able to test it him/herself), a random sample of 30 new calls revealed a mean length of time the advertisement ran before the listeners hung-up was 2.3 minutes with a standard deviation of 0.8 minutes. To test the claim that the mean length of time the advertisement is heard with music is greater than the mean time without the music, it was assumed that the distributions of ‘listening’ times for both scenarios were approximately normally distributed with unequal population variances. The null hypothesis is Ho:________ and the alternative hypothesis is H1:________, with the claim located in ________ (fill in this blank with either Ho or with H1 as appropriate). The appropriate distribution to use to test this claim is the ________ (chi-square, F, t, z-standard normal) and using a 0.05 significance level the critical value or values obtained from that distribution is/are ___________ (this blank should be completed with the appropriate numerical value or values of the critical value/s). Furthermore, the test statistic value is calculated to be __________ (complete this blank with the appropriate numerical value of the test statistic), which was computed using _________ (fill in this blank with the appropriate selection from the list of formulas below.)
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b) page 427:
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Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Of primary concern is how much of a pre-recorded 7-minute advertisement is heard. In a random sample of 16 calls a mean of 1.85 minutes of the ad was ‘heard’ with a standard deviation of 1.4 minutes.
To test this employee’s claim that listening time would increase, a random sample of 30 new calls revealed a mean length of time before the listeners hung-up was 2.3 minutes with a standard deviation of 0.8 minutes.
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To test the claim that the mean length of time the advertisement is heard with music is greater than the mean time without the music, it was assumed that the distributions of ‘listening’ times for both scenarios were approximately normally distributed with unequal population variances.
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The null hypothesis is
Ho: u2-u1 <= 0
H1:u2-u1 > 0 (claim)
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The appropriate distribution to use to test this claim is the t distribution
and using a 0.05 significance level the critical value or values obtained from that distribution is/are (df = 15, t = +-invT(0.025 when df = 15 = +- 2.131 (this blank should be completed with the appropriate numerical value or values of the critical value/s).
Furthermore, the test statistic value is calculated to be
t = [2.3-1.85]/sqrt[1.4^2/16)+(2.3^2/30) = 0.8231 (complete this blank with the appropriate numerical value of the test statistic
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Conclusion: Since the test statistic is not in either reject region,
fail to reject Ho. The test result does not support the claim.
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Cheers,
Stan H.
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