e XYZ Company operates service centers where customers can call to get answers t

Question

e XYZ Company operates service centers where customers can call to get answers to questions about their bills. Previous orúdies indicate that the distribution of time required for each call is normally distributed with a mean 540 seconds. Company officials have selected a random sample of 36 calls and wish to determine whether the mean call time is now fewer,than 540 seconds after a training program is given to call-center employees ate the hypotheses associated with this research. Determine the p-value if the sample mean is found to be 520 seconds and the sample standard deviation is 45 seconds. Ho = 540 / HA =

Explanation / Answer

Part a

The null and alternative hypotheses for the given scenario or test is given as below:

Null hypothesis: H0: The mean call time to get answers to questions is 540 seconds.

Alternative hypothesis: Ha: The mean call time to get answers to questions is less than 540 seconds.

H0: µ = 540 versus Ha: µ < 540

Part b

Here, we have to use one sample t test for the population mean.

H0: µ = 540 versus Ha: µ < 540

We are given

Xbar = 520

S = 45

n = 36

df = n – 1 = 36 – 1 = 35

= 0.01

Test statistic formula is given as below:

t = (Xbar - µ) / [ S / sqrt(n) ]

t = (520 – 540) / [45/sqrt(36)]

t = -20 / [45/6 ]

t = -20/ 7.5

t = -2.6667

Lower critical value = -2.4377

(By using t-table)

P-value = 0.0058

(By using t-table)

= 0.01

P-value < = 0.01

So, we reject the null hypothesis that the mean call time to get answers to questions is 540 seconds.

There is sufficient evidence to conclude that the mean call time to get answers to questions is less than 540 seconds.

Part c

For the given scenario, the maximum should be 0.0058 for which the null hypothesis would not be rejected.

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.