Question

Quarterly returns on an index are approximately normally distributed with a standard deviation of 10%. Index returns over the most recent 20 quarters have a mean of 4%. Should a researcher who wants to show that the average quarterly returns for this index are positive reject the null hypothesis at the 5% significance level?

Answer 1

A	B	C	D	E	F
2		Sample size	20
3		Sample Mean	4.0%
4		Sample St. Deviation	10.0%
5		T-test needs to be performed.
6		Inputs--
7		Hypothesized Population mean (µ):	0.0%
8		Sample Standard Deviation (s):	10.0%
9		Sample Size (n):	20
10		Sample mean (X-bar):	4.00%
11		Alpha	5.0%
12		Hypothesis formulation:
13		Null hyposthesis is an assertion that is true unless there is sufficient statistical evidance to conclude otherwise.
14		Alternation hypothesis is the negation of the null hypothesis.
15
16		Null Hypothesis: µ > 0%
17		Alternate Hypothesis: µ <= 0%.
18		H0: µ > 0%
19		H1: µ <= 0%
20
21		Intermediate calculation--
22		Standard Error of the Estimate(S.E.):	0.02	=s/sqrt(n)
23		Test Statistic(t):	1.79	=(X-bar - µ)/S.E.
24		Degree of freedom (d.f.):	19	=n-1
25
26		Results:
27		One tailed, p value, P(T>t)	0.045	=T.DIST.RT(D23,D24)
28
29
30		Alpha	0.05	(for 5% significance level)
31		Null hypothesis should be	Rejected	(As p-value is lower than alpha)
32
33		Thus it can be concluded that for a significance level of 5%, the null hypothesis should be rejected.
34

Formula sheet

A	B	C	D	E	F
2		Sample size	20
3		Sample Mean	0.04
4		Sample St. Deviation	0.1
5		T-test needs to be performed.
6		Inputs--
7		Hypothesized Population mean (µ):	0
8		Sample Standard Deviation (s):	=D4
9		Sample Size (n):	20
10		Sample mean (X-bar):	=D3
11		Alpha	0.05
12		Hypothesis formulation:
13		Null hyposthesis is an assertion that is true unless there is sufficient statistical evidance to conclude otherwise.
14		Alternation hypothesis is the negation of the null hypothesis.
15
16		Null Hypothesis: µ > 0%
17		Alternate Hypothesis: µ <= 0%.
18		H0: µ > 0%
19		H1: µ <= 0%
20
21		Intermediate calculation--
22		Standard Error of the Estimate(S.E.):	=D8/SQRT(D9)	=s/sqrt(n)
23		Test Statistic(t):	=(D10-D7)/D22	=(X-bar - µ)/S.E.
24		Degree of freedom (d.f.):	=D9-1	=n-1
25
26		Results:
27		One tailed, p value, P(T>t)	=T.DIST.RT(D23,D24)	=getformula(D27)
28
29
30		Alpha	=D11	(for 5% significance level)
31		Null hypothesis should be	Rejected	(As p-value is lower than alpha)
32
33		Thus it can be concluded that for a significance level of 5%, the null hypothesis should be rejected.
34