Respuesta :
Answer:
The p-value of the test is 0.0901 < 0.1, which means that this provides evidence at the 10% significance level that the company has reached its goal of decreasing the percentage of complaints.
Step-by-step explanation:
A company hopes to improve customer satisfaction, setting a goal of less than 5% negative comments.
At the null hypothesis, we test if the proportion of negative comments is of at least 5%, that is:
[tex]H_0: p \geq 0.05[/tex]
At the alternative hypothesis, we test if this proportion is less than 0.05, that is:
[tex]H_1: p < 0.05[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.05 is tested at the null hypothesis:
This means that [tex]\mu = 0.05, \sigma = \sqrt{0.05*0.95}[/tex]
A random survey of 850 customers found only 34 with complaints.
This means that [tex]n = 850, X = \frac{34}{850} = 0.04[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.04 - 0.05}{\frac{\sqrt{0.05*0.95}}{\sqrt{850}}}[/tex]
[tex]z = -1.34[/tex]
P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion below 0.04, which is the p-value of z = -1.34.
Looking at the z-table, z = -1.34 has a p-value of 0.0901.
The p-value of the test is 0.0901 < 0.1, which means that this provides evidence at the 10% significance level that the company has reached its goal of decreasing the percentage of complaints.
