Question

A large national survey shows that the waiting times for fast food restaurants is approximately normally distributed with a mean of μ = 3.4 minutes and a standard deviation of σ = 0.6 minutes. A fast food outlet claims that their mean waiting time in line is less than 3.4 minutes. To test the claim, a random sample of 60 customers ia selected, and is found to have a mean of 3.3 minutes Use this sample data to test the fast food outlet's claim at a significance leve of α = 0.05. Show all 7 steps for p-value method.

Answer 1

Solution :

Given that ,

= 3.4

= 3.3

= 0.6

n = 60

Stap - 1. The null and alternative hypothesis is ,

H₀ :   = 3.4

H_a : < 3.4

2. This is the left tailed test .

Test statistic = z

= ( - ) / / n

= ( 3.3 - 3.4) / 0.6 / 60

= -1.29

3. The test statistic = -1.29

P - value = P (Z < -1.29 ) = 0.0985

4. P-value = 0.0985

= 0.05  

0.0985 > 0.05

P-value >

5. The p - value ie greater than

6. Fail to reject the null hypothesis .

7. Conclusion :- Fail to reject the null hypothesis .There is not sufficient evidence to test the claim.