Question

A random sample of size n= 15 obtained from a population that is normally distributed results in a sample mean of 45.8 and sample standard deviation 12.2. An independent sample of size n = 20 obtained from a population that is normally distributed results in a sample mean of 51.9 and sample standard deviation 14.6. Does this constitute sufficient evidence to conclude that the population means differ at the a = 0.05 level of significance? Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Click here to view the table of critical t-values. Click here to view the chi-square critical values table. The given situation is about a Write the hypotheses for the te mean, . HO: proportion, p. H: standard deviation, o, or variance o2. Calculate the test statistic. (Round to two decimal places as needed.) Identify the critical region. Select the correct choice below and fill in all answer boxes within your choice. (Type an integer or decimal rounded to two decimal places as needed.) O A. test statistic OB. test statistic > OC. test statistics or test statistic > What is the conclusion? the null hypothesis and conclude there sufficient evidence that the parameter of interest of population 1 is the parameter of interest of population 2 at the a = 0.05 level of significance.

Answer 1

Q1:

The given situation is about a mean µ.

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 32.5533 = 33

Test statistic:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (45.8 - 51.9)/√(12.2²/15 + 14.6²/20) = -1.34

Critical value, t_c = T.INV.2T(0.05, 33) = 2.03

Reject Ho if t < -2.03 or if t > 2.03

Conclusion:

Fail to reject the null hypothesis and conclude there is not sufficient evidence that the parameter of interest of population 1 is different than the parameter of interest of population 2 at the 0.05 level of significance.

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Q2:

a)

B. A test regarding the difference of two means using Welch's approximate t.

b)

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

c)

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 56.4802 = 56

Test statistic:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (123.3 - 129.8)/√(8.5²/31 + 7.3²/50) = -3.5270

p-value :

p-value =T.DIST.2T(-3.527, 56) = 0.001

Decision:

A. There is sufficient evidence to reject the null hypothesis because the P-value < alpha.