Question

Fowle Marketing Research, Inc., bases charges to a client on the assumption that telephone surveys can be completed in a mean time of 15 minutes or less. If a longer mean survey time is necessary, a premium rate is charged. Suppose a sample of 35 surveys produces the data in the Microsoft Excel Online file below. Use a known σ = 3.3 minutes. Is the premium rate justified?

Call Time (minutes)

20.3
13.7
16.2
18.1
13.5
17.3
14.6
17.4
18.5
9.9
15.1
16.6
17
14.5
19.7
11.3
16.3
20.5
21.7
15.9
12.2
18.5
25.9
14.5
21.6
15.6
15.9
14.7
17.6
18
16.7
11.6
16.8
13
13.2

Hypothesized Mean value 15

Population Standard Deviation 3.3

Level of Significance 0.01

1. Formulate the null and alternative hypotheses for this application.

   H₀: μ _________greater than or equal to 15greater than 15less than or equal to 15less than 15equal to 15not equal to 15

   H_a: μ _________greater than or equal to 15greater than 15less than or equal to 15less than 15equal to 15not equal to 15

2. Compute the value of the test statistic (to 2 decimals).

3. What is the p-value (to 4 decimals)?

4. Using ? = .01, is a premium rate justified for this client?

Answer 1

Solution:

Given:

Hypothesized Mean value = 15

Population Standard Deviation = 3.3

Level of Significance = 0.01

If a longer mean survey time is necessary, a premium rate is charged. That is if mean survey time is > 15 minutes, then   premium rate is charged.

Part a) Formulate the null and alternative hypotheses for this application.

Vs

Part b) Compute the value of the test statistic

where

For given sample , total of all observations is 573.9

that is: and n = 35

thus

thus z test statistic is:

Part c) What is the p-value ?

p-value = P( Z > z test statistic value)

p-value = P( Z > 2.50 )

p-value = 1 - P( Z < 2.50 )

Look in z table for z = 2.5 and 0.00 and find corresponding area.

P( Z < 2.50 ) = 0.9938

thus

p-value = 1 - P( Z < 2.50 )

p-value = 1 - 0.9938

p-value = 0.0062

Part d) Using ? = .01, is a premium rate justified for this client?

Since p-value = 0.0062 < 0.01 level of significance , we reject null hypothesis H0.

Thus we have enough evidence to conclude that: mean survey time is more than 15 minutes and hence premium rate is charged.

Thus a premium rate is NOT justified for this client.

Thus answer is: No.