Answer:
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.
