Question

In: Statistics and Probability

You recently started working as a nurse manager at a larger local urgent care facility. Because...

You recently started working as a nurse manager at a larger local urgent care facility. Because the center is so new (and a bit fancier than most) many patients come here expecting a bit of a wait before they can be seen. Out of curiosity, on a particularly busy day you write down how many minutes a random sample of patients has to wait before being seen by one of the medical staff. Using a significance level of 20%, test whether the average wait time is less than 15 minutes.

12	10	15	17	15	18	15	15
10	15	12	10	12	19	10	14
10	17	13	15	23	12	15	20
7	10	16	14	14	14	18	13

A) What is the name of the type of hypothesis test required here? Is this a left-tailed, right-tailed, or two-tailed test? B) State AND verify all assumptions required for this test. [HINT: this test has two assumptions.] C) State the null and alternate hypotheses for this test: (use correct symbols and format!) Null hypothesis/Alternate hypothesis D) Run the correct test in Excel or on the TI-84 calculator and provide the information below. For each statistic provide the correct symbol and numeric value (round answers to 3 decimal places).Degrees of Freedom. Test Statistic.Critical Value. p-value E) State your statistical decision for this test and justify it using the statistics above. F) Interpret your decision within the context of the problem: what is your conclusion?

Expert Solution

A)

Hypothesis test for one mean - t test

left tail test

b)

assumptions: random independent samples are used

population from which samples are taken is normally distributed

c)

Ho : µ = 15
Ha : µ < 15

d)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   3.4355
Sample Size ,   n =    32
Sample Mean,    x̅ = ΣX/n =    14.0625

degree of freedom=   DF=n-1=   31

Standard Error , SE = s/√n =   3.43546494   / √    32   =   0.6073
t-test statistic= (x̅ - µ )/SE = (   14.063   -   15   ) /    0.607   =   -1.544

critical t value, t* = -0.8534 [Excel formula =t.inv(α/no. of tails,df) ]
p-Value = 0.0664 [Excel formula =t.dist(t-stat,df) ]

e)
Decision: p-value<α, Reject null hypothesis

f)

Conclusion: There is enough evidence to conlcude that average wait time is less than 15 minutes at 20% level of significance

orchestra answered 2 years ago

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