Question

1) Use the Student-t distribution and the sample results to complete the test of
the hypotheses. Use a 5% significance level. Assume the results come from a
random sample, and if the sample size small, assume the underlying
distribution is relatively small.
a) Test ?!: ? = 10 ?? ?!: ? > 10 using the sample results ? = 13.2 , ? =
8.7 with ? = 12
b) Test ?!: ? = 120 ?? ?!: ? < 120 using the sample results ? = 112.3 , ? =
18.4 with ? = 100
c) Test ?!: ? = 4 ?? ?!: ? ≠ 4 using the sample results ? = 4.8 , ? = 2.3 with
? = 15

2) A t-test for a mean uses a sample of 15 observations. Find the t-test statistic
value that has a P-value of 0.05 when the alternative hypothesis is
a) ?!: ? ≠ 0
b) ?!: ? > 0
c) ?!: ? < 0

3) A study has a random sample of 20 subjects. The t-test statistic for testing
?!: ? = 100 is ? = 2.40. Find the approximate P-vale for alternative
a) ?!: ? ≠ 100
b) ?!: ? > 100
c) ?!: ? < 100

Answer 1

1)a) The test statistic t = ()/(s/)

                                    = (13.2 - 10)/(8.7/)

                                    = 1.274

P-value = P(T > 1.274)

             = 1 - P(T < 1.274)

             = 1 - 0.8855

             = 0.1145

Since the P-value is greater than the significance level(0.1145 > 0.05), so we should not reject the null hypothesis.

b) The test statistic t = ()/(s/)

                                 = (112.3 - 120)/(18.4/)

                                 = -4.18

P-value = P(T < -4.18)

             = 0.000

Since the P-value is greater than the significance level(0 < 0.05), so we should reject the null hypothesis.

c) The test statistic t = ()/(s/)

                                 = (4.8 - 4)/(2.3/)

                                 = 1.347

P-value = 2 * P(T > 1.347)

             = 2 * (1 - P(T < 1.347))

             = 2 * (1 - 0.9003)

             = 0.1994

2) a) 2 * P(T < t) = 0.05

        or, P(T < t) = 0.025

        or, t = -2.145

b) P(T > t) = 0.05

or, 1 - P(T < t) = 0.05

or, P(T < t) = 0.95

or, t = 1.761

c) P(T < t) = 0.05

or, t = -1.761

3)a) P-value = 2 * P(T > 2.4)

                 = 2 * (1 - P(T < 2.4))

                 = 2 * (1 - 0.9866)

                 = 0.0268

b) P-value = P(T > 2.4)

                 = 1 - P(T < 2.4)

                 = 1 - 0.9866

                 = 0.0134

c) P-value = P(T < 2.4)

                 = 0.9866