Question

Test the hypothesis that the average weekly allowance of female and male ACC 215 students the same, at 5% level of sign...


Test the hypothesis that the average weekly allowance of female and male ACC 215 students the same, at 5%
level of significance.


Weekly Allowance of Students (Php)

Male: 1500 2000 2500 2500 2000 1750 2000 1800
Female:1500 2000 1750 1500 2000 2500 3000 3500 

Ho:
Ha:
α =
Decision Rule:
Computation:


Decision:

Conclusion: 

Answer

To compare the means of two populations with different standard deviations the t-test with unequal variance is used. For this test pooled standard deviation is not used.

The test statistic for this test is computed by t=x¯1-x¯2s12n1+s22n2, where x¯1-x¯2 is the difference in sample means, s1, s2 are sample standard deviation and n1, n2 area sample sizes. The degree of freedom for this test is calculated using df=s12n1+s22n22(s12n1)2n1-1+(s22n2)2n2-1.

It is asked to test whether the weekly mean allowance is equal or different for male and female. So the null hypothesis will claim that the mean allowance is same and alternate hypothesis will claim that mean allowance is not equal.

H0:μ1=μ2Ha:μ1μ2

Compute the sample mean and sample standard deviation by applying the formula AVERAGE() and STDEV.S(), respectively in excel. This gives the following results.

x¯1=2006.25s1=348.914x¯2=2218.75s2=725.031

The standard deviation will be considered different if the ratio of higher standard deviation to smaller standard deviation is more than 2.

s2s1=725.031348.9142.1

The ratio of standard deviation is more than 2 so it can be considered that the population standard deviations are different and thus t-test with unequal variance can be used.

The level of significance is 5% or 0.05 which is α so α=0.05. The alternate hypothesis has symbol  so the test is two tailed. Critical value will be equal to the t value corresponding to two tail area 0.05.

Compute the degree of freedom for the test by substituting s1=348.914, s2=725.031n1=8 and n2=8 in df=s12n1+s22n22(s12n1)2n1-1+(s22n2)2n2-1.

df=348.91428+725.03128348.9142828-1+725.0312828-1=6549072390649887191.710

Using the table of t distribution critical value corresponding to two tail area 0.05 and 10 degrees of freedom is 2.228.

Decision rule: If the test statistic lies outside from -2.228 to 2.228 then the null hypothesis can be rejected.

Compute the test statistic by substituting x¯1=2006.25, x¯2=2218.75, s1=348.914, s2=725.031, n1=8 and n2=8 in t=x¯1-x¯2s12n1+s22n2.

t=2006.25-2218.75348.91428+725.03128=-212.5284.4756-0.747

So the test statistic is -0.747.

Decision: The test statistic lies between -2.228 and 2.228 so the null hypothesis cannot be rejected.

Conclusion: The null hypothesis cannot be rejected so it can be said that the average allowance for male and female is equal.

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