2. A partner of a walk-in dental dlinic suspected that the mean waiting time for

Question

2. A partner of a walk-in dental dlinic suspected that the mean waiting time for patients had increased from the past average of 40 minutes. The partner sampled 18 patients and recorded their waiting times. She wanted to see whether the data supported her belief. Here is a stem-and-leaf plot of the 18 times: 2 79 3 1446 4 01234789 5 56 6 12 Note: 413= 43 minutes (4 points) Can we answer this question using the t-procedure for a hypothesis test? Explain your reasoning? a. b. (12 points) REGARDLESS of your response in part a, conduct a test to determine if the average waiting time has increased. Use a 10% level of significance. (Note, the above data produces the values x = 43.3 and s 10.56)

Explanation / Answer

SolutionA:

n=18

since sample size=n=18

n<30

and population standard deviation is not given,we can use the t procedure for a hypothesis test.

Solutionb:

Null hypothesis:

Ho: =40

Alternative Hypothesis:

H1: >40

Level of significance=10%=0.1

test statistic:

given sample:

27,29,31,34,34,36,40,41,42,43,44,47,48,49,55,56,61,62

sample mean=779/18=43.3

sd=10.56

t=samplemean-popmean/samplesd/sqrt(n)

=43.3-40/10.56/sqrt(18)

=1.326

t cal=1.326

df=n-1=18-1=17

alpah/2=0.1/2=0.05

t crit=1.740

T cal<t crit

1.326<1.740

Fail to reject Null hypothesis.
Accept Null hypothesis.

There is no sufficient statistical evidence at 10% level of signiifcance to conclude that the average waiting time has increased from 40 minutes.

sample(x) xbar x-xbar (x-xbar)^2 27 43.3 -16.3 265.69 29 43.3 -14.3 204.49 31 43.3 -12.3 151.29 34 43.3 -9.3 86.49 34 43.3 -9.3 86.49 36 43.3 -7.3 53.29 40 43.3 -3.3 10.89 41 43.3 -2.3 5.29 42 43.3 -1.3 1.69 43 43.3 -0.3 0.09 44 43.3 0.7 0.49 47 43.3 3.7 13.69 48 43.3 4.7 22.09 49 43.3 5.7 32.49 55 43.3 11.7 136.89 56 43.3 12.7 161.29 61 43.3 17.7 313.29 62 43.3 18.7 349.69 total 1895.62 Sample sd=sqrt(x-xbar)^2/n-1      =SQRT(1895.62/18-1) samplesd 10.21334 samplesd=10.21334

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