Question

2. (20) Fo r α-001, find the test statistic, critical value, P-value, and statistical deckion for the following questions: (a) H1 : μ 69, ®--67.6, s-3, and n-24. (b) Hi : p < 0.4, p = 0.37 and n-1021. (c) HI : μι 7,42両= 69.3, 쪼2 = 68.5, σ1 = σ2 = 3,m = n2= 16. (d) Hi : μ1关μ2,峦1 = 12.2両= 11.5, si = 0.00, s2 = 0.65, n.-n-12, and s,-osa.

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 69
Alternative hypothesis: u $\neq$ 69

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.6124

DF = n - 1

D.F = 23
t = (x - u) / SE

t = - 2.29

t_critical = + 2.808

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 23 degrees of freedom is less than -2.29 or greater than 2.29.

Thus, the P-value = 0.032

Interpret results. Since the P-value (0.032) is less than the significance level (0.05), we have to reject the null hypothesis.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.40
Alternative hypothesis: P < 0.40

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.01533
z = (p - P) / S.D

z = - 1.956

z_critical = - 2.327

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than -1.956

Thus, the P-value = 0.0252

Interpret results. Since the P-value (0.0252) is greater than the significance level (0.01), we failed to reject the null hypothesis.

c)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁$\neq$ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (t).

SE = sqrt[($\sigma$₁²/n₁) + ($\sigma$₂²/n₂)]
SE = 1.06066
z = [ (x₁ - x₂) - d ] / SE

z = 0.754

z_critical = + 2.576

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a z statistic more extreme than -0.754; that is, less than -0.754 or greater than 0.754.

Thus, the P-value = 0.451

Interpret results. Since the P-value (0.451) is greater than the significance level (0.01), we failed to reject the null hypothesis.

d)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁$\neq$ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s_p(sqrt[(1/n₁) + (1/n₂)])
SE = 0.25311
DF = 22
t = [ (x₁ - x₂) - d ] / SE

t = 2.77

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 22 degrees of freedom is more extreme than -2.77; that is, less than -2.77 or greater than 2.77.

Thus, the P-value = 0.011

Interpret results. Since the P-value (0.011) is greater than the significance level (0.01), we failed to reject the null hypothesis.