Understanding this hypothesis testing exercise using P-value The exercise states that:

Your team evaluates a new cannon ball and compares its distance
performance to a new ball from your competitor. In an initial
evaluation you test 10 of your new cannon balls and 10 of the
competitor’s balls with the goal of determining if the average
distance in yards travelled by your new ball is greater than the
average distance travelled by the competitor’s ball. The resulting
data is shown below. You might assume equal variances when using a
statistical test. Which one of the following statements is correct?

*

*The properly calculated P-value is in the range 0.15 < P <0.45. You have statistical evidence that your new ball travels farther on
average than the competition’s ball.


*None of the statistical hypothesis tests discussed are appropriate
for the analysis of this data.


*The properly calculated P-value is in the range 0.075 < P< 0.25. No evidence is found that your new ball travels farther on average
than the competition’s new ball.


*A statistical hypothesis test cannot be applied in this case, because we have less than 12 data points for each type of ball.

 




Competitor ball
My new ball




341.0
314.2


308.2
315.0


314.0
354.2


306.4
301.2


291.7
327.8


333.1
332.0


314.2
320.5


318.0
314.6


301.6
321.3


319.0
311.1




I'm using Minitab, Two-Sample t for the Mean for this exercise. My question is about what the right interpretation is for the alternative hypothesis. It could be u1-u2 ≠ 0 or u1-u2 < 0 which means u2 (my ball's mean) is greater than u1 (competitor's ball mean).
If u1-u2 ≠ 0 then P value = 0.330 and the answer is option (1).
If u1-u2 < 0, then P value = 0.165 but this does not really fit the options.



 A: I put your data into R. Both Welch and pooled 2-sample t tests
gave $T\approx -1,$ and P-value about $0.168 > 5\%$ for a one-sided test
that your cannon balls go farther than your competitor's.
This is in agreement with your first block of Minitab output
and with item (3) in your multiple-choice list.
I think the question is poorly worded, but decipherable, and don't
understand why you did not choose response (3). Please leave a note
if you need further clarification.

DTA = matrix(c(
341.0,  314.2,
308.2,  315.0,
314.0,  354.2,
306.4,  301.2,
291.7,  327.8,
333.1,  332.0,
314.2,  320.5,
318.0,  314.6,
301.6,  321.3,
319.0,  311.1), byrow=T, ncol=2)

t.test(DTA[,1], DTA[,2], alt="less")

        Welch Two Sample t-test

data:  DTA[, 1] and DTA[, 2]
t = -1.0017, df = 18, p-value = 0.1649
alternative hypothesis: 
  true difference in means is less than 0
95 percent confidence interval:
     -Inf 4.730536
sample estimates:
 mean of x mean of y 
    314.72    321.19 

Boxplots and stripcharts show variability too large to distinguish
a significant difference in sample locations.
boxplot(DTA, col="skyblue2", pch=20, horizontal=T)


stripchart(list(DTA[,1], DTA[,2]), pch="|", ylim=c(.5,2.5))


