Question

Life on the prairie is hard on soldiers in 1876, and so a LOT of drinking occurs. General George Armstrong "Autie" Custer wants to know whether there is a significant difference in the shooting ability of his soldiers when sober vs. when in their cups. He invites the soldiers to a shooting competition and collects data on their target shooting.   The numbers represent how many times a man-size target was hit in 20 attempts by people in each group.

Soldier                 Sober Soldier                    After a few

    Joe                            10                                 8

    Fred                          12                                 6

    Tom                          1                               1

    Zebediah                   17                            14

    Henry                       6                                8

    Caleb                        5                               5

    Japheth                    20                              14

    Phillip                        6                                 8

    Jeremy                      16                               9

a. Find the mean difference.

b. Find s²_D  (variance term).

c. Find sD  (standard error).

d. Find t.

e. Find df (degrees of freedom).

   g. With a t critical value of 2.306, what do you conclude about Ho?

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).

a) Dbar = 2.2222 = Mean difference

b)

s² = [ (\sum (d_i - d)² / (n - 1) ]

s² = 12.1945

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

s = 3.4921

SE = s / sqrt(n)

S.E = 1.164

DF = n - 1 = 9 -1

D.F = 8

t = [ (x₁ - x₂) - D ] / SE

t = 1.91

tcritical = +2.036

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.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 8 degrees of freedom is more extreme than 1.91; that is, less than - 1.91 or greater than 1.91.

Thus, the P-value = 0.093

Interpret results. Since the P-value (0.093) is more than the significance level (0.05), we have to accept the null hypothesis.

Do not reject H0. The mean difference appears to be zero.