Question

In testing whether the means of two normal populations are equal, summary statistics computed for two independent samples are as follows:

Brand X Brand Y

𝑛1 = 25 𝑛₂ = 25

x=7.3 x=6.8

S₁=1.05 S₂=1.20

Assume that the population variances are equal. Then the 95% confidence interval for the difference between population means of band 𝑋 and brand 𝑌 is:

1. (-0.1407; 1.1407)

2. (0.5; 0.0906)

3. (0.490; 0.9506)

4. (0.3976; 0.5609)

5. (0.5906; 0.4094)

Answer 1

Solution:

$𝑛_1 = 25, 𝑛_2 = 25 \\\bar x_1=7.3, \bar x_2=6.8 \\S_1=1.05, S_2=1.20$

$H_0:\mu_1= \mu_2 \\ H_1:\mu_1\ne \mu_2$

α = 0.05      

Degrees of freedom = 25 + 25 - 2 = 48

Lower Critical t- score = -2.011     

Upper Critical t- score = 2.011

Now, we reject $H_0$ if $|t|>2.011$

Pooled variance, $S_p=\sqrt{\dfrac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}}$

$S_p=\sqrt{\dfrac{(24)(1.05)^2+(24)(1.2)^2}{48}}=1.127$

Then, SE$=S_p\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}=1.127\sqrt{\dfrac{2}{25}}=0.318$

Now, 95% CI$=(\bar x_1-\bar x_2)\pm t_{0.05,48}\ SE$

$=(7.3-6.8)\pm2.011(0.318) \\=(-0.139,1.139) \\\approx(-0.14, 1.14)$

Thus, option 1 is correct.