Question

A lead-free solder can be made using two different alloys. An important characteristic

feature of a solder is its melting point, which is known to follow a Normal

distribution. An experiment was conducted using a random sample of 21 pieces of

solder made from the two different alloys (1 and 2). In each sample, the temperature

at which each of the 21 pieces melted was determined. The mean and standard

deviation of the sample for Alloy 1 were M1= 218.9ºC and S1 = 2.7ºC; for Alloy 2 the

results were M2 = 215.5ºC and S2 = 3.6ºC. If it is required to test H0: µ1 = µ2 against

HA: µ1 ≠ µ2, where µ stands for the population mean, what would be the value of the

test statistic?

Answer 1

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$$F = \frac{s_1^2}{s_2^2} = \frac{ 2.7^2}{ 3.6^2} = 0.563$$

The critical values are $F_L = 0.406$ and $F_U = 2.464,$ and since $F_L=0.406\leq F = 0.563\leq F_U=2.464,$ then the null hypothesis of equal variances is not rejected.

The degrees of freedom are computed as follows, assuming that the population variances are equal: $df=n_1-1+n_2-1=40.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{M_1-M_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_2}+\dfrac{1}{n_2})}}$$

$$=\dfrac{218.9-215.5}{\sqrt{\dfrac{(21-1)(2.7)^2+(21-1)(3.6)^2}{21+21-2}(\dfrac{1}{21}+\dfrac{1}{21})}}$$

$$\approx3.4624$$

The value of the test statistic is $3.4624.$