A blanching process currently used in the canning industry consists of treating vegetables with a large volume of boiling water before canning. A newly developed method, called Steam Blanching Process (SBP), is expected to remove less vitamins and minerals from vegetables, because it is more of a steam wash than a flowing water wash. Ten batches of string beans from different farms are to be used to compare the SBP and the standard process. One-half of each batch of beans is treated with the standard process; the other half of each batch is treated with the SBP. Measurements of the vitamin content per pound of canned beans are:
SBP Standard
35 33
48 40
65 55
33 41
61 62
54 54
49 40
37 35
58 59
65 56
a) Do the data provide strong evidence that SBP removes less vitamins in canned beans than the standard method of blanching? Use t-test and sign-test.
b) Construct a $98$% confidence interval for the difference between the mean vitamin contents per pound using the two methods of blanching.
Attempted Solution
a)
t-test
$H_0$: $\delta$ = $0$
$H_1$: $\delta$ $\gt$ $0$
Average difference $\bar{D}$ = $3$
$S_D$ = $5.8689$
t = $\frac{\bar{D}-\delta_0}{S_D\over{\sqrt{n}}}$ = $1.616$
$t_{9, .05}$ = $1.83313$
Since $1.616 \lt 1.833113$ we fail to reject $H_0$ at $\alpha = .05$
Sign Test
Proportion of positive $D_i$'s = $2\over3$
$P(X \geq 6)$ = $.5^{9}$[$9\choose{6}$ + $9\choose{7}$ + $9\choose{8}$ + $9\choose{9}$]
= $.254 \gt .05$ so we fail to reject $H_0$ at $\alpha = .05$
b) $\bar{D}$ $\pm$ $t_{n-1, \alpha/2}$$\frac{S_D}{\sqrt{n}}$
=$3$ $\pm$ $t_{9, .01}$$\frac{5.8689}{\sqrt{10}}$
= $(-2.236, 8.236)$
Did I do these correctly? Particularly the sign test?