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.

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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?

Solutions

Expert Solution

Hypothesis test for one mean - t test

left tail test

assumptions: random independent samples are used

population from which samples are taken is normally distributed

Ho : µ = 15
Ha : µ < 15

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

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

orchestra answered 1 year ago

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Subjects

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

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

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