More hypothesis help needed in statistics?

Question

A restaurant claims their service time is less than 18 minutes. A random sample of 36 customers resulted in a mean service time of 17.1 minutes with a standard deviation of 3.1 minutes. Is there enough evidence to support a claim alpha=0.08 Perform an appropriate hypothesis test showing each important step.

Answer 1

M's answer is wrong. Gab23's answer is definitively correct.

Answer 2

ANSWER: "true mean" of restaurant service times are to a 0.08 level of significance greater than 18 minutes.

SINGLE SAMPLE TEST, ONE-TAILED, 7 - Step Procedure for t Distributions, "one-tailed test"

1. Parameter of interest: "μ" = population mean fax restaurant service times in minutes.

2. Null hypothesis Ho: μ < 18 minutes

3. Alternative hypothesis Ha: μ ≥ 18 minutes

4. Test statistic formula: t = (x-bar - μ)/(s/SQRT(n))

x-bar = estimate of the Population Mean (statistical mean of the sample) [17.1]

n = number of individuals in the sample [36]

s = sample standard deviation [3.1]

μ = Population Mean [18] (used for Test statistic)

5. Computation of Test statistic formula t = -1.74

6. Determination of the P-value: The test is based on n -1 = 35 df (degrees of freedom). Table "look-up" value shows area under the 35 df curve to the left of t = -1.74 is (approximately) 0.045

7. Conclusion: with significance value α = 0.08 the above shows P-value <= α, [0.045 <= 0.08]. Null hypothesis Ho: μ = 18 should be rejected. "true mean" of restaurant service times are to a 0.08 level of significance greater than 18 minutes.

Answer 3

EDITED.

I think the above answerer is wrong, We use the z score, NOT the t score in this case.

The question requires us to test that

(Null hypothesis) Ho: U=18. (the restaurant's mean service time is equal to 18 minutes).

i.e.,U=18.

versus

(Alternative hypothesis) Ha: U<18.(the restaurant's mean service time is less than 18 minutes).

i.e., U<18.

According to the central limit theorem, since the given sample size of n=36 is large(at least greater than 30), then the sampling distribution of the restaurant's mean service time(_x)- the probability distribution of all possible sample means of the restaurant's service time obtained from all possible samples of size n= 36 would be approximately normally distributed and hence it is appropriate to describe and estimate the above probability distribution using the z distribution in this case. So we can estimate or replace the true population standard deviation of the restaurant's service time- δ by its sample standard deviation- s.

With this particular sample which has size of n=36, _x= sample mean =17.1 and s = sample standard deviation=3.1,

so the test static

= z

= (_x-U0)/(s/√n)

= (17.1-18)/(3.1/√36)

= -1.74

Given that the school's enrolment officer requires us to set and test at α =0.08 level of significance, then

the rejection point = -zα

= -z0.08

= -1.405

The p-value

= The area under the standard normal curve to the left of z

= P(z>0) - P(-1.74<z<0)

= 0.5 - 0.4591

= 0.0409

Rejection criteria:

We can only reject Ho in favor of Ha if and only if z<-zα or if the p-value is less than that our chosen value of α.(which is 0.08 in this case).

Conclusion: Because z of -1.74 is less than the rejection point (-zα=-z0.08 =-1.405), then we can reject Ho in favor of Ha at α =0.08 level of significance. So we conclude that the sample data does support the claim that the restaurant's mean service time is less than 18 minutes. In other words, since the p value of 0.0409 is less than α=0.08, so the restaurant has reasonable evidence to support its claim, based on its collected set of sample data.

Hope this helps.