Question

A random sample of size n1=25,n_1 = 25, taken from a normal population with a standard deviation σ1=5.2\sigma_{1}=5.2 has a mean x1=81.\overline{x}_{1}=81. A second random sample of size n2=36,n_{2}=36, taken from a different normal population with a standard deviation σ2=3.4,\sigma_{2}=3.4, has a mean x2=76\overline{x}_{2}=76 Test the hypothesis that μ1=μ2\mu_{1}=\mu_{2} against the alternative, μ1μ2.\mu_{1} \neq \mu_{2}. Quote a P-value in your conclusion.

Solution

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We have:

n1=25,σ1=5.2,x1=81n_{1}=25, \\ \sigma_{1}=5.2, \\ \overline{x}_{1}=81

n2=36,σ2=3.4,x2=76n_{2}=36, \\ \sigma_{2}=3.4, \\ \overline{x}_{2}=76

The null hypothesis is:

H0:μ1=μ2H_{0} : \mu_{1}=\mu_{2}

The alternative hypothesis is:

H1:μ1μ2H_{1} : \mu_{1} \neq \mu_{2}

Therefore, we have

H0:μ1μ2=0H_{0} : \mu_{1}-\mu_{2}=0

And

H1:μ1μ20H_{1} : \mu_{1}-\mu_{2} \neq 0

We will use the level of significance α=0.05\alpha=0.05

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