Question

4 Friday the 6th: Friday the 13th: 9 13 5 11 11 14 11 10 6 10 What are the hypotheses for this test? Let Hd be the in the numbers of hospital admissions resulting from motor vehicle crashes for the population of all pairs of data Ho Po 0 Hi Ho VO Find the value of the test statistic =(Round to three decimal places as needed.) Identify the critical value(s). Select the correct choice below and fill the answer box within your choice. (Round to three decimal places as needed.) O A. The critical values are t = 1 OB. The critical value is t= State the result of the test. Choose the correct answer below. O A. There is sufficient evidence to warrant rejection of the claim of no effect Hospital admissions do not appear to be affected. OB. There is not sufficient evidence to warrant rejection of the claim of no effect. Hospital admissions do not appear to be affected. O C. There is sufficient evidence to warrant rejection of the claim of no effect. Hospital admissions appear to be affected. OD. There is not sufficient evidence to warrant rejection of the claim of no effect. Hospital admissions appear to be affected

Researchers collected data on the numbers of hospital admissions resulting from motor vehicle crashes, and results are given below for Fridays on the 6th of a month and Fridays on the following 13th of the same month. Use a 0.05 significance level to test the claim that when the 13th day of a month falls on a friday, the number of hospital admissions for motor vehicle crashes are not affected.

Answer 1

the necessary calculation table:-

friday 6 th (xi)	friday 13 the (yi)	dị = Ti-gi.	(P - !p)
9	13	-4	0.444489
5	11	-6	7.111289
11	14	-3	0.111089
11	10	1	18.77749
4	6	-2	1.777689
4	10	-6	7.111289
		sum=-20	sum=35.333333

sample size (n) = 6

Σd, d -20 6 = -3.3333 η

Sd Σ(d, - d)2 η –1 35.333333 6 - 1 & 2.6583

a).let
Pr
be the difference in the .........................

hypothesis:-

[ claim ]

HO: Ud=0

Hi :d +0

b).the test statistic be:-

t= d Sdn -3.3333 2.6583/V6 -3.071

c). df = (n-1) = (6-1) = 5

t critical value for df = 5, alpha=0.05, both tailed test = 2.571

the critical values are t = $\pm$ 2.571 (A)

d).There is sufficient evidence to warrant rejection of the claim of no effect. Hospital admissions appear to be affected . (C)

[ t calculated = -3.071 < -2.571...so, we reject the null hypothesis .]

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