Question

The melting points of two alloys used in formulating solder were investigated by melting 21 samples of each material. The sample mean and standard deviation for alloy 1 was 420.48 and 2.34 respectively, and for alloy 2 they were 425 and 2.5 respectively. Suppose that the true mean difference in melting points is 3 F. How large a sample would be required to detect this difference using an alpha = 0.05 level test with probability at least 0.9? Use sigma = 4 as an initial estimate of the common standard deviation. Use at least 2 decimal digits

Answer 1

calculate abscissa scale factor

d |ܙܠ - ܠܐ܂ 3 - 0| 2(4) 0,375 ?c

n^* from operating characteristic curve for
g= 0,05
with
d= 0.375
and $\beta = 0.1$

n = 75

and

$n=\frac{n^*+1}{2}=\frac{75+1}{2}=\frac{76}{2}=38$