Question

A standard painkiller used for patients after minor surgeries is known to bring relief in 3.5 minutes on the average (μ). 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 α = 0.05.

The test statistic is closest to:

Answer 1

The test statistic is closest to: -2.7738

Solution:

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

H0: µ = 3.5 versus Ha: µ < 3.5

This is a lower tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 3.5

Xbar = 2.8

S = 1.1

n = 19

df = n – 1 = 18

α = 0.05

Critical value = -1.7341

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (2.8 - 3.5)/[1.1/sqrt(19)]

t = -2.7738

P-value = 0.0063

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the new painkiller indeed works faster.

The data provide sufficient evidence to indicate that the new painkiller indeed works faster.