Question

14. (10 pts.) A crayon manufacturer is comparing the effects of two kinds of yellow dye on the brittleness of crayons. Dye B is more expensive than dye A, but it is thought that it might produce a stronger crayon. Four crayons are tested with each kind of dye, and the impact strength (in joules) is measured for each. The results are as follows: | Dye A | 1.5 Dye B 3.2 | 2.0 3.1 | 1.1 3.1 2.2 | 2.0 2.4 | 3.7 | Use a hypothesis test to determine if mean strength of crayons made with dye B is greater than that of crayons made with dye A.

Answer 1

Solution:

Let $\mu$ A and $\mu$ B be the mean strength of crayons made with dye A and dye B respectively.

To test if the mean strength of crayons made with dye B is greater than dye A.

1)Null and alternative hypothesis.

H0:$\mu$B -$\mu$A$\leq$0

H1:$\mu$B -$\mu$A>0

2)I have used excel to solve test statistics.

In excel choose Data->data analysis->t-test:Paired two sample for means

D 1.5 1 Dye A 2 Dye B C 2 3.1 1.1 2 E 3.1 2.4 F 2.2 3.7 3.2 - Data Analysis ? x 5 t-Test: Paired Two Sample fo OK Cancel Analysis Tools Exponential Smoothing F-Test Two-Sample for Variances Fourier Analysis Histogram Moving Average Random Number Generation Rank and Percentile Regression Sampling t-Test: Paired Two Sample for Means Help 8 Mean 9 Variance 10 Observations 11 Pearson Correlation 12 Hypothesized Mean Differer 13 df 14 t Stat 15 PIT<=t) one-tail 16 lt Critical one-tail 17 P(T<=t) two-tail 18 t Critical two-tail -2.12 0.05 2.13 0.10 2.78

The output is as follows:

	Dye A	Dye B
Mean	1.98	2.88
Variance	0.577	0.457
Observations	5	5
Pearson Correlation	0.130475448
Hypothesized Mean Difference	0
df	4
t Stat	-2.12
P(T<=t) one-tail	0.0506
t Critical one-tail	2.13
P(T<=t) two-tail	0.10
t Critical two-tail	2.78

From the above table we can see that P-value=0.0506 which is greater than $\alpha$ =0.05, we fail to reject the null hypothesis.

So we can conclude that the mean strength of crayons made with dye B is not greater than dye A.