Question

A USA based cable broadband company wants to compare the average customer satisfaction scores between its east coast and west coast customer bases. The customer survey asks for a score between 1 and 5, with 1 being poor and 5 being excellent.174 east coast customers are surveyed, and the sample mean is 3.51 with a sample standard deviation of 0.51. For the west coast, 355 customers are surveyed, and the sample mean is 3.24 with a sample standard deviation of 0.52.a. Create a 95% confidence interval for the average score for west coast customers. Interpret this interval in context of the problem.b. Now say you consider the difference between the average scores between west coast and east coast. Create a 95% confidence interval for the difference in average scores (specify what difference you are considering).c. The company wants to investigate if the average score of the east coast customers is higher than the average score of the west coast customers. Write the null and alternative to test this hypothesis.d. Compute the test statistic for this problem and describe how you would obtain the p-value for this test.e. Say the p-value is very close to 0. In a sentence or two explain what that means in context of this problem.

Answer 1

a. The 95% confidence interval for the average score of west coast customers is (3.17, 3.30). b. The 95% confidence interval for the difference in average scores between west coast and east coast customers is (-0.16, -0.07). c. The null hypothesis is that the average score of the east coast customers is equal to the average score of the west coast customers, and the alternative hypothesis is that the average score of the east coast customers is higher than the average score of the west coast customers.

Step-by-step explanation:

a. To create a 95% confidence interval for the average score of west coast customers, we can use the formula:

Confidence Interval = sample mean ± (critical value * sample standard deviation / sqrt(sample size))

Using the given information, the confidence interval for the average score of west coast customers is 3.24 ± (1.96 * 0.52 / sqrt(355)), which simplifies to (3.17, 3.30). This means that we are 95% confident that the true average score for west coast customers falls within this range.

b. To create a 95% confidence interval for the difference in average scores between west coast and east coast customers, we can use the formula:

Confidence Interval = (sample mean of west coast - sample mean of east coast) ± (critical value * sqrt((sample standard deviation of west coast)^2 / sample size of west coast + (sample standard deviation of east coast)^2 / sample size of east coast))

Using the given information, the confidence interval for the difference in average scores is (3.24 - 3.51) ± (1.96 * sqrt((0.52)^2 / 355 + (0.51)^2 / 174)), which simplifies to (-0.16, -0.07). This means that we are 95% confident that the true difference in average scores between west coast and east coast customers falls within this range.

c. The null hypothesis for this test is that the average score of the east coast customers is equal to the average score of the west coast customers. The alternative hypothesis is that the average score of the east coast customers is higher than the average score of the west coast customers.

d. To compute the test statistic, we can use the formula:

Test Statistic = (sample mean of east coast - sample mean of west coast) / sqrt((sample standard deviation of east coast)^2 / sample size of east coast + (sample standard deviation of west coast)^2 / sample size of west coast)

In this case, the test statistic is (3.51 - 3.24) / sqrt((0.51)^2 / 174 + (0.52)^2 / 355), which simplifies to 4.804. The p-value can be obtained by comparing the test statistic to the critical value(s) associated with the desired level of significance. For example, if we are using a significance level of 0.05, we would compare the test statistic to the critical value(s) for a t-distribution with the appropriate degrees of freedom.

e. If the p-value is very close to 0, it means that the observed difference in average scores between east coast and west coast customers is unlikely to have occurred by chance alone, assuming the null hypothesis is true. In other words, there is strong evidence to suggest that the average score of the east coast customers is higher than the average score of the west coast customers.

Answer 2

Step-by-step explanation:

a) For west coast customers, 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.

Margin of error = z × s/√n

Where

s = sample standard deviation = 0.52

n = number of samples = 174

From the information given, the population standard deviation is unknown and the sample size is small, 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 = 174 - 1 = 173

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 = 1.9738

Margin of error = 1.9738 × 0.52/√174

= 0.078

the lower limit of this confidence interval is

3.24 - 0.078 = 3.162

the upper limit of this confidence interval is

3.24 + 0.078 = 3.318

Therefore, we are 95% confident that the population mean score for west coast customers is between 3.162 and 3.318

b) Confidence interval for the difference in the two population means is written as

Difference in sample mean ± margin of error

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = sample mean of east coast customers

x2 = sample mean of west coast customers

s1 = sample standard deviation of east coast customers

s2 = sample standard deviation of west coast customers

n1 = number of east coast customers surveyed

n2 = number of west coast customers surveyed

For a 95% confidence interval, the z score is 1.96

From the information given,

x1 = 3.51

s1 = 0.51

n1 = 174

x2 = 3.24

s2 = 0.52

n2 = 355

x1 - x2 = 3.51 - 3.24 = 0.27

Margin of error = 1.96√(0.51²/174 + 0.52²/355) = 1.96√0.00225651773

= 0.093

Confidence interval = 0.27 ± 0.093

c) This is a test of 2 independent groups. Let μ1 be the mean score of east coast customers and μ2 be the mean score of wet coast customers.

The random variable is μ1 - μ2 = difference in the mean score between east coast and west coast customers

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 ≥ μ2 H0 : μ1 - μ2 ≥ 0

The alternative hypothesis is

H1 : μ1 < μ2 H1 : μ1 - μ2 < 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (3.51 - 3.24)/√(0.51²/174 + 0.52²/355)

t = 5.68

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.51²/174 + 0.52²/355]²/[(1/174 - 1)(0.51²/174)² + (1/355 - 1)(0.52²/355)²] = 0.00000509187/0.00000001456

df = 350

We would determine the probability value from the t test calculator. It becomes

p value < 0.00001

This is very close to 0

Since the p value is very small, there is a very high possibility of rejecting the null hypothesis. It means that the difference in the mean score between east coast and west coast customers is less than zero. Therefore, the average score of the east coast customers is not higher than the average score of the west coast