A consultant working for a State Police barracks contends that service weapons w
ID: 3224151 • Letter: A
Question
A consultant working for a State Police barracks contends that service weapons will fire with a higher muzzle velocity if the barrel properly cleaned. A random sample of Glock 9-mm handgun was obtained, and the muzzle velocity (in feet per second) of single shot from each gun was measured. Each gun was professionally cleaned, and the muzzle velocity of a second shot (with the same bullet type) was measured. The data are given in the following table GLOCK a. What is the common characteristic that makes these data paired? b. Assume normality. Conduct the appropriate hypothesis test to determine whether there is any evidence that a clean gun fires with a higher muzzle velocity. Use alpha = 0.01. c. Find bounds on the p value associated with this hypothesis test.Explanation / Answer
a. The data sets are "paired" because the following one-to-one relationship exists between values in the two data sets.
1 . Each data set has the same number of data points.
2. Each data point in one data set (Before) is related to one, and only one, data point in the other data set (After).
b. Null Hypothesis H0: The average velocity of the gun after cleaned is equal to the average velocity of the gun before cleaned
Alternative Hypothesis H1: The average velocity of the gun after cleaned is greater than the average velocity of the gun before cleaned.
The difference of the average speed of pairs is 120 92 -45 -53 -38 15
The mean of difference d = 15.167
Standard deviation of the difference, SD = 74.815
Standard error, SE = SD/sqrt(n) = 74.815/sqrt(6) = 30.5431
Degree of freedom = n-1 = 6-1 = 5
t = 15.167/30.5431 = 0.4966
p-value for t = 0.4966 at df = 5 , is 0.3203
As, p-value is greater than the significance level of 0.01, we fail to reject the null hypothesis and there is not sufficient statistical evidence to prove that the average velocity of the gun after cleaned is greater than the average velocity of the gun before cleaned.
(c) t = 0.4966
From the t-table, we get
0 < 0.4966 < 0.690
0.5 > P(T > 0.4966) > 0.25
As this is one-tail paired test, we can reasonably say that 0.25 < P-value < 0.5
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