Question

In: Statistics and Probability

Use the following information: n_1=35,x ̅_1=5,s_1=3,n_2=32,x ̅_2=7,s_2=5 Test whether the score variance for the first population...

Use the following information: n_1=35,x ̅_1=5,s_1=3,n_2=32,x ̅_2=7,s_2=5 Test whether the score variance for the first population is significantly different from that of the second population at the significant level α=0.05, to do so: State null and alternative hypotheses. Compute the test-value. Find the critical value. Make your decision and explain the reason. Use the result part (a) for assumption on variance of two populations, test whether the second population mean is greater the first one at the level α=0.05, to do so state, The list of assumptions. Null and alternative hypotheses. Compute the test-value. Find the critical value. Make your decision and explain the reason.

Solutions

Expert Solution

Solution:

First we have to use F test for difference in population variances.

H0: The score variance for the first population is not significantly different from that of the second population.

Ha: The score variance for the first population is significantly different from that of the second population.

H0: σ12 = σ22 versus H0: σ12 ≠ σ22

We are given α = 0.05

S1^2=3^2=9,

S2^2=5*5=25

n1 = 35

n2 = 32

Test statistic = F = S2^2/S1^2 = 25/9 = 2.7778

P-value = 0.0043 (by using F-table)

(by using F-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the score variance for the first population is significantly different from that of the second population at the significant level α=0.05.

Now, we have to use two sample t test for population means.

H0: Both population means are equal.

Ha: Second population mean is greater than first population mean.

H0: µ1 = µ2 vs. Ha: µ1 < µ2

Two sample t test for mean assuming unequal population variances

t = (X1bar – X2bar) / sqrt[(S12 / n1)+(S22 / n2)]

t = (5 – 7) / sqrt[(3^2/35)+(5^2/32)]

t = -1.9627

df = n1 + n2 – 2 = 35+32-2 = 85

P-value = 0.0265 (by using t-table)

P-value < α=0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the score variance for the first population is significantly different from that of the second population at the significant level α=0.05.


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