Question

4. For random variable X with unknown variance let S_x^2=49,x ̄=8,n=25,"and" α=0.10. Conduct the following tests. Make sure you state the decision rule and the inference.

a) H_0:μ<7H_A:μ≥7

b) H_0:μ>10H_A:μ≤10

c) H_0:μ=3H_A:μ≠3

d) Give a 90% confidence interval around .

Answer 1

a)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 7
Alternative Hypothesis: μ > 7

Rejection Region
This is right tailed test, for α = 0.1 and df = 24
Critical value of t is 1.318.
Hence reject H0 if t > 1.318

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (8 - 7)/(9/sqrt(25))
t = 0.556

P-value Approach
P-value = 0.292
As P-value >= 0.1, fail to reject null hypothesis.

b)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 10
Alternative Hypothesis: μ < 10

Rejection Region
This is left tailed test, for α = 0.1 and df = 24
Critical value of t is -1.318.
Hence reject H0 if t < -1.318

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (8 - 10)/(9/sqrt(25))
t = -1.111

P-value Approach
P-value = 0.139
As P-value >= 0.1, fail to reject null hypothesis.

c)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 3
Alternative Hypothesis: μ ≠ 3

Rejection Region
This is two tailed test, for α = 0.1 and df = 24
Critical value of t are -1.711 and 1.711.
Hence reject H0 if t < -1.711 or t > 1.711

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (8 - 3)/(9/sqrt(25))
t = 2.778

P-value Approach
P-value = 0.01
As P-value < 0.1, reject the null hypothesis.

d)

sample mean, xbar = 8
sample standard deviation, s = 7
sample size, n = 25
degrees of freedom, df = n - 1 = 24

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.71

ME = tc * s/sqrt(n)
ME = 1.71 * 7/sqrt(25)
ME = 2.394

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (8 - 1.71 * 7/sqrt(25) , 8 + 1.71 * 7/sqrt(25))
CI = (5.606 , 10.394)