Question

In: Statistics and Probability

A company hopes to improve customer satisfaction, setting as a goal no more than 5% negative...

A company hopes to improve customer satisfaction, setting as a goal no more than 5% negative comments. A random survey of 350 customers found only 10 with complaints. Does this provide evidence that the company has reached its goal of getting less than 5% negative comments?

a) What is the sample statistic? Use correct notation. Round to three decimal places.

b) Write the null and alternative hypothesis. (Hint: The null is 5% or 0.05. Also, think carefully about the inequality in your alternative.)

c) Find the test statistic, z. Round to two decimal places. Show your work.

d) What is the p-value? Use Statkey or your TI-84 calculator to find the p-value. Round to three decimal places if necessary.

e) At a 5% significance level, will you reject or fail to reject H0?

f) Write your conclusion in context.

Solutions

Expert Solution

Solution:

Here, we have to use one sample z test for the population proportion.

a) What is the sample statistic?

We have

x = number of items of interest = 10

n = sample size = 350

Sample statistic = p̂ = x/n = 10/350 = 0.029

b) Write the null and alternative hypothesis.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: The Company has reached its goal of getting not less than 5% negative comments.

Alternative hypothesis: Ha: The Company has reached its goal of getting less than 5% negative comments.

H0: p ≥ 0.05 versus Ha: p < 0.05

This is a lower tailed test.

We are given

Level of significance = α = 0.05

c) Find the test statistic, z.

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

x = number of items of interest = 10

n = sample size = 350

p̂ = x/n = 10/350 = 0.028571429

p = 0.05

q = 1 - p = 1 - 0.05 = 0.95

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.028571429 - 0.05) / sqrt(0.05*0.95/350)

Z = -1.8394

Test statistic = -1.84

d) What is the p-value?

P-value = 0.033

(by using Ti-84 calculator)

e) At a 5% significance level, will you reject or fail to reject H0?

P-value < α = 0.05

So, we reject the null hypothesis

f) Write your conclusion in context.

There is sufficient evidence to conclude that The Company has reached its goal of getting less than 5% negative comments.


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