Question

1b)

A standard painkiller used for patients after minor surgeries is known to bring relief in 3.5 minutes on the average (u). A new painkiller is hypothesized to bring faster relief to patients. To test this hypothesis, a sample of 19 patients with minor surgeries are selected and given the new painkillers. This sample yields a mean of 2.8 minutes and a standard deviation of 1.1 minutes. Do the data provide sufficient evidence to indicate that the new painkiller indeed works faster? Assume the times to relief are normally distributed and use a = 0.05. Which of the following pairs of hypotheses is appropriate for this study? O Ho: M = 3.5 vs. Hail< 3.5 O Ho:ll = 2.8 vs. Ha:M < 2.8 O Ho: M = 3.5 vs. Ha:M < 2.8 O Ho: M = 3.5 vs. Ha:ll #3.5 O Ho: M = 3.5 vs. Ha:ll> 3.5

The test statistic is closest to:

Group of answer choices:

a) t = -0.64

b) t = 0.64

c) t = 2.77

d) t = -2.77

e) t = 3.5

1c) The rejection region for the test is closest to:

Group of answer choices:

a) (-\infty ∞ , -1.729] \cup ∪ [1.729, \infty ∞ )

b) (-\infty ∞ , -1.645] \cup ∪ [1.645, \infty ∞ )

c) (-\infty ∞ , -1.645]

d) (-\infty ∞ , -1.734]

e) (-\infty ∞ , -1.729]

1d) The correct decision and justification are:

Group of answer choices:

a) Do not reject H0 because the p-value is greater than \alpha α = 0.05

b) Do not reject H0 because the test statistic does not fall in the rejection region

c) Reject H0 because the test statistic falls in the rejection region

d) Reject H0 because z < -1.645.

e) Reject H0 because 2.8 < 3.5.

Answer 1

Given that 2.8,s 1.1,n = 19 Null and the alternative hypothesis: H u3.5 0 H u3.5 Significance level, a = 0.05 Test statistic 2.8-3.5 1.1 V19 =-2.77 Degree of freedom: n-1 = 19-1 =18 Critical value By using t tables, the critical value of t at % level of significance for right tailed test is, toos =1.734 Decision rule: Reject Ho if t <-t or do not reject H The rejection region for the test is [-,-1.734] Do not reject HO because the test statistic does not fall in the rejection region