Question

Question 1 (30 points): Two different types of glass bottles, type 1 and type 2, are suitable for use by a soft-drink beverage bottler. The internal pressure strength of the bottle is an important quality characteristic. From two random samples of the two bottle types of sizes n1 = n2 = 15 bottles, the average pressure strengths of the type 1 and 2 samples are observed to be 183 and 177 psi, respectively, and the standard deviations for the two samples are 2.5 and 2 psi, respectively. The company will use bottle design 1 if its pressure strength exceeds that of bottle design 2 by at least 5 psi, otherwise, they will use bottle design 2. Using a = 0.05: a) Based on the collected samples, and assuming that the two bottle designs have the same variances in internal pressure strength, should they use bottle design 1 or 2? (15 Points) b) If type 1 sample contained 6 defect bottles, and type 2 sample contained 1 defected bottle, can you conclude that both types have different proportions of defected bottles? What is the p-value for your test? (15 Points)

Answer 1

given that

for bottle 1

n1=15 sample mean =m1=183 sample SD=S1=2.5

for bottle 2

n2=15 sample mean =m2=177 sample SD=S2=2

a)

we nee to test that bottle 1 mean is at least 5 psi more than botthle 2 mean or not hence

$H_0: \mu_1-\mu_2 \geq 5,H_a: \mu_1-\mu_2 < 5$

as we have assumed that population SDs are equal so we will use pooled SD which is given as below

$S=\sqrt{\frac{(n1-1)S1^2+(n2-1)S2^2}{n1+n2-2}}=\sqrt{\frac{(15-1)*2.5^2+(15-1)*2^2}{15+15-2}}=2.26$

now test statistics is given by

$t=\frac{m_1-m_2-5}{S*\sqrt{\frac{1}{n1}+\frac{1}{n2}}}=\frac{183-177-5}{2.26*\sqrt{\frac{1}{15}+\frac{1}{15}}}=1.21$

t have DF=(n1+n2-2) =(15+15-2) =28

now test is Left tailed so

P-Value =P(t<1.21) =0.88

Since P-value is more than level of significance hence failed to reject H0 as we dont have enough evidence to support the claim that mean of bottle 1 is less than 5 more than that of bottle 2

so we will use bottle 1

b)

since

n1=n2=15

number of defected in sample 1 =x1=6

number of defected in sample 2=x2=15

so sample proportions are

$\hat{P}_1=\frac{x_1}{n_1}=\frac{6}{15}=0.4,\hat{P}_2=\frac{x_2}{n_2}=\frac{1}{15}=0.07$

we have to test that

$H_0: P_1-P_2=0,H_1: P_1-P_2 \neq 0$

now calculating pooled proportion

$P=\frac{x_1+x_2}{n_1+n_2}=\frac{6+1}{15+15}=0.23$

now test statistics is given by

$Z=\frac{\hat{P}_1-\hat{P}_2}{\sqrt{P*(1-P)*(\frac{1}{n_1}+\frac{1}{n_2})}}=\frac{0.4-0.07}{\sqrt{0.23*(1-0.23)*(\frac{1}{15}+\frac{1}{15})}}=2.15$

since test is two tailed so

P-Value =2*P(Z>2.15) =2*0.02=0.04

since P value is less than level of significance hence we reject H0 that is there is enough evidence to support the claim that there is difference in the proportions.