contestada

The manager of a grocery store took a random sample of 100 customers. The avg. length of time it took the customers in the sample to check out was 3.1 minutes with a std. deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly > 3 min. At 95% confidence, it can be concluded that the mean of the population is

Respuesta :

ajeigbeibraheem

Answer:

Step-by-step explanation:

The data given are;

sample size n = 100

sample mean x = 3.1

standard deviation σ = 0.5

mean = 3

The value for Z can be determined by using the formula:

[tex]Z = \dfrac{x - \mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

[tex]Z = \dfrac{3.1 - 3.00}{\dfrac{0.5}{\sqrt{100}}}[/tex]

[tex]Z = \dfrac{0.1}{\dfrac{0.5}{10}}}[/tex]

Z = 0.2

At 95% Confidence interval, level of significance ∝ = 0.05

From the z table ;P- value for the test statistics at ∝ = 0.05

P = 0.0228

We can see that the P-value is < ∝

Decision Rule:

Reject the null hypothesis [tex]H_o[/tex] if P-value is less than ∝

Conclusion:

At 0.05 level of significance; we conclude that the mean of the population is significantly > 3 min