Answer:
a) No sufficient evidence to support the claim that the two machines produce rods with different mean diameters.
b) P-value is 0.80
c) −0.3939 <μ< 0.4939
Step-by-step explanation:
Given Data:
sample sizes
n1 = 15
n2 = 17
sample means:
x1 = 8.73
x2 = 8.68
sample variances:
s1² = 0.35
s2² = 0.40
Hypothesis:
H₀ : μ₁ = μ₂
H₁ : μ₁ ≠ μ₂
Compute the pooled standard deviation:
[tex]s_{p} = \sqrt{\frac{(n_{1} - 1)s_{1}^{2} + (n_{2} - 1)s_{2}^{2}}{n_{1} +n_{2} -2} }[/tex]
[tex]= \sqrt{\frac{(15-1)0.35+(17-1)0.40}{15+7-2}}[/tex]
[tex]= \sqrt{\frac{(14)0.35+(16)0.40}{30}}[/tex]
[tex]= \sqrt{\frac{4.9+6.4}{30}}[/tex]
[tex]= \sqrt{\frac{11.3}{30}}[/tex]
[tex]= \sqrt{0.376667}[/tex]
= 0.613732
= 0.6137
Compute the test statistic:
[tex]t = \frac{x_{1} -x_{2} }{s_{p} \sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } } }[/tex]
[tex]= \frac{8.73-8.68}{0.6137\sqrt{\frac{1}{15}+\frac{1}{17} } }[/tex]
[tex]= \frac{0.05}{0.6137\sqrt{0.06667+0.05882} } }[/tex]
[tex]= \frac{0.05}{0.6137\sqrt{0.12549} } }[/tex]
[tex]= \frac{0.05}{0.6137(0.354246)} } }[/tex]
[tex]= \frac{0.05}{0.6137(0.354246)} } }[/tex]
= 0.05 / 0.217401
= 0.22999
t = 0.230
Compute degree of freedom:
df = n1 + n2 -2 = 15 + 17 - 2 = 30
Compute the P-value from table using df = 30
P > 2 * 0.40 = 0.80
P > 0.05 ⇒ Fail to reject H₀
Null hypothesis is rejected when P-value is less than or equals to level of significance. But here the P-value = 0.80 and level of significance = 0.05. So P-value is greater than significance level. Hence there is not sufficient evidence to support the claim that population means are different.
Construct a 95% confidence interval for the difference in mean rod diameter:
confidence = c = 95% = 0.95
α = 1 - c
= 1 - 0.95
α = 0.05
Compute degree of freedom:
df = n1 + n2 -2 = 15 + 17 - 2 = 30
Compute [tex]t_{\alpha /2}[/tex] with df = 30 using table:
t₀.₀₂₅ = 2.042
Compute confidence interval:
= [tex](x_{1}-x_{2})-t_{\alpha/2} ( s_{p} )\sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } }[/tex]
= (8.73 - 8.68) - 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]
= 0.05 - 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]
= 0.05 - 1.253175 [tex]\sqrt{0.06667+0.05882} } }[/tex]
= 0.05 - 1.253175 [tex]\sqrt{0.12549} } }[/tex]
= 0.05 - 1.253175 (0.35424))
= 0.05 - 0.443925
= −0.393925
= −0.3939
[tex](x_{1}-x_{2})+t_{\alpha/2} ( s_{p} )\sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } }[/tex]
= (8.73 - 8.68) + 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]
= 0.05 + 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]
= 0.05 + 1.253175 [tex]\sqrt{0.06667+0.05882} } }[/tex]
= 0.05 + 1.253175 [tex]\sqrt{0.12549} } }[/tex]
= 0.05 + 1.253175 (0.35424))
= 0.05 + 0.443925
= 0.493925
= 0.4939
−0.3939 <μ₁ - μ₂< 0.4939
