Question 847524
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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