An office manager has implemented an incentive plan that she thinks will reduce the mean time required to handle a customer complaint. The mean time for handling a complaint was 30 minutes prior to implementing the incentive plan. After the plan was in place for several months, a random sample of the records of 38 customers who had complaints revealed a mean time of 28.7 minutes with a standard deviation of 3.0 minutes.(a) Give a point estimate of the mean time required to handle a customer complaint. 28.7(b) What is the standard deviation of the point estimate given in (a)? 0.49(c) Construct a 95% confidence on the mean time to handle a complaint after implementing the plan. (27.451, 29.949)Interpret the confidence interval for the office manager.(d) Is there sufficient evidence that the incentive plan has reduced the mean time to handle a complaint? (Use ? = .05.)I NEED P VALUE AND Z VALUE HELPP value=z =

Step-by-step explanation:

a)The point estimate is the same as the sample mean. Therefore, the point estimate of the mean time required to handle a customer complaint is 28.7 minutes

b) The standard deviation of the point estimate given in (a) is 3 minutes

c) Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Since the sample size is large and the population standard deviation is known, we would use the following formula and determine the z score from the normal distribution table.

Margin of error = z × σ/√n

Where

s = sample standard deviation = 3

n = number of samples = 38

From the information given, the population standard deviation is unknown hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 38 - 1 = 37

Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 2.0262

Margin of error = 2.0262 × 3/√38

= 0.99

the lower limit of this confidence interval is

28.7 - 0.99 = 27.71

the upper limit of this confidence interval is

28.7 + 0.99 = 29.69

Therefore, we are 95% confident that the true mean time for handling a complaint is between 27.71 minutes and 29.69 minutes

d) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 150

For the alternative hypothesis,

µ < 30

This is a left tailed test

Since the population standard deviation is not given, the distribution is a student's t.

Degrees of freedom, df = n - 1 = 38 - 1 = 37

t = (x - µ)/(s/√n)

Where

x = sample mean = 28.7

µ = population mean = 30

s = samples standard deviation = 3

t = (28.7 - 30)/(3/√38) = - 2.67

We would determine the p value using the t test calculator. It becomes

p = 0.0056

Since alpha, 0.05 > than the p value, 0.0056, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed sufficient evidence that the incentive plan has reduced the mean time to handle a complaint.

use google you idiot omg you're so annoying

step-by-step explanation:

quit

step-by-step explanation: