Question

6. Paired annual rates of return data are collected from 8 randomly selected investment funds before and after The Federal Reserve cuts down interest rates. The dataset and relevant summery results are given in the table below. Suppose you are a financial analyst interested in finding out whether investment funds’ mean rate of return is significantly different before and after the interest rate adjustment.

Fund	Before (%)	After (%)	After-Before (%)
1	3.51	4.62	1.11
2	4.25	4.31	0.06
3	1.76	1.52	-0.24
4	2.68	2.69	0.01
5	3.19	3.77	0.58
6	5.43	4.86	-0.57
7	2.18	3.69	1.51
8	6.72	7.98	1.26
Average=	3.72	4.18	0.47
Standard Deviation=	1.68	1.88	0.76

The ALTURNATIVE hypothesis of this test is ________________________________________.

The significance level for this test should be chosen to be _______________________.

The numerical formula calculating test statistic is __________________________________________.

The test statistic is calculated to be_________________________.

The p-value is ___________________________.

Based on the p-value we _________________, (accept or reject H0)

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).

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

s = 0.76463

SE = s / sqrt(n)

S.E = 0.2703

DF = n - 1 = 8 -1

D.F = 7

t = [ (x₁ - x₂) - D ] / SE

t = -1.72

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 7 degrees of freedom is more extreme than -1.72; that is, less than - 1.72 or greater than 1.72.

Thus, the P-value = 0.129.

Interpret results. Since the P-value (0.129) is greater than the significance level (0.05), we have to accept the null hypothesis.

Do not reject H₀.