Answer to Question #267850 in Statistics and Probability for Adona

Question

1. Audio cassette tapes of a particular brand are claimed by the manufacturer to give, on average, at least 60 minutes of playing time.  After receiving some complaints, the manufacturer’s quality control manager obtains a random sample of 64 tapes and measures the playing time, t minutes, of each.  The results are summarized by

 t=3953.28 and t²=244,557.00

a. Constructing a symmetric 99% confidence interval for the population mean playing time of the brand of tape.

b. Does the confidence interval support the customer’s complaints? Give a reason for your answer.

Answer 1

a)

average of sample:

$\overline{x}=3953.28/64=61.77$

$z=\frac{\overline{x}-\mu}{\sigma/\sqrt n}$

$\sigma=\sqrt{t^2/(n-1)}=\sqrt{244557/63}=62$

critical values for 99% confidence interval:

$z=\pm 2.576$

$-2.576<\frac{61.77-\mu}{62/\sqrt{64}}<2.576$

$-19.96<61.77-\mu<19.96$

$41.81<\mu<81.73$

b.

The confidence interval support the customer’s complaints, because mean playing time can be less than 60 minutes (minimal value is 41.81 min).