The waiting time for customers at macburger restaurants follows a normal distribution with a population standard deviation of 1 minute. at the warren road macburger, the quality assurance department sampled 50 customers and found that the mean waiting time was 2.75 minutes. at the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?
(a) state the null hypothesis and the alternate hypothesis.
(b) state the decision rule.
(c) compute the value of the test statistic.
(d) what is your decision regarding h0?
(e) what is the p-value? interpret it.

Step-by-step explanation:

a)

Null hypothesis (H0): μ=3 min

Alternative hypothesis (H1): μ<3 min

b)

The decision rule is:

t-statistic< T (t-student table) --> You must accept the null hypothesis

t-statistic > T (t-student table) --> You must reject the null hypothesis

c) t-statistic formula:

t= (ybar-m)/(σ/(sqrt(n)))

ybar: sample mean

m: hypothesized value

σ: sample standard deviation

n: number of observations

t=(2.75-3)/(1/sqrt(50))

t= -1.76

d)

The t-student table statistic at 5% significance level and for a sample of 50 observations is: 1.6759. Because this test only uses the lower tail the statisctic value is: -1.6759

-1.76>-1.6759

Then, you must reject the null hypothesis and accept the alternative hypothesis: the mean waiting time is less than 3 minutes.

e)

You must use a tail probability table for the t-student distribution with a significance level o 5% and n-1 freedem degrees, in this case 49. Also you can use the excel formula "NORMSINV"

Accodding to the excel formula, the p-value is: 0,039

Note that with this result we also reject the null hypothesis because the p-value is less than the significance level: 0.03<0.05

You can conclude that the mean waiting time is less than 3 minutes.

24+x=y

step-by-step explanation: