Question

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 falils on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected. Friday the 6th: 101046 12 Friday the 13th: What are the hypotheses for this test? Let ud be the 14 14 9 614 in the numbers of hospital admissions resulting from motor vehicle crashes for the population of all pairs of data. 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 fillthe answer box within your choice. (Round to three decimal places as needed.) O A. The critical value is t OB. The critical values are ?. State the result of the test. Choose the correct answer below A. There is not sufficient evidence to warrant rejection of the claim of no effect Hospital admissos do ? B. There is sufficient evidence to warrant rejection of te dann or no efect. Hospital admissions spear is be aneea O C. There is not sufficient evidence to warrant rejection of the claim of no effect. Hospital admissions appear to be affected Click to select your answer(s). pe here to search Envd PrtScn FT

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d = 0

Alternative hypothesis: u_d ? 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 3.4303

SE = s / sqrt(n)

S.E = 1.4004

DF = n - 1 = 6 -1

D.F = 5

t = [ (x₁ - x₂) - D ] / SE

t = - 2.74

t_critical = + 2.571

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Interpret results. Since the t-value (-2.74) is less than the t-value (- 2.571), we have to reject the null hypothesis.

Reject H0. The mean difference appears to differ from zero.

(B) There is sufficient evidence to warrant rejection of the claim of no effect. Hospitals admissions appear to be affected.