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babymother [125]
4 years ago
12

A consumer agency is interested in examining whether there is a difference in two common sealant products used to waterproof res

idential backyard decks. With cooperation of several builders in the area, they randomly assign 38 newly constructed decks to be treated with Very Clear deck sealant and another 37 newly constructed decks to be treated with Sure Seal deck sealant. After one year of being exposed to similar weather conditions, the decks are rated on a scale of 1 to 100. The mean rating for the decks treated with Very Clear is 89.2 with a standard deviation of 3.1. The mean rating for the decks treated with Sure Seal is 92.4 with a standard deviation of 3.8.
Which of the following represents the 90 percent confidence interval to estimate the difference (Very Clear minus Sure Seal) in mean ratings for the two deck sealants?

a.(89.2−92.4)±1.9603.1238+3.8237−−−−−−−−√
b.(89.2−92.4)±1.6883.1238+3.8237−−−−−−−−√
c.(89.2−92.4)±1.6453.1238+3.8237−−−−−−−−√
d.(89.2−92.4)±1.6883.138+3.837−−−−−−−√
e.(89.2−92.4)±1.645(3.138√+3.837√)
Mathematics
1 answer:
Komok [63]4 years ago
8 0

Answer:

CI = (89.2 - 92.4) \pm 1.6883 \sqrt{\frac{3.1^{2} }{38} +\frac{3.8^{2} }{37} }

Step-by-step explanation:

Data of newly constructed decks treated with very clear deck sealants:

n₁ = 38, \bar {x_{1} } = 89.2,  S₁ = 3.1

Data of newly constructed decks treated with sure seal deck sealants:

n₁ = 37, \bar {x_{2} } = 92.4, S₂ = 3.8

First calculate the degree of freedom

df = \frac{[\frac{s_{1} ^{2} }{n_{1} }  + \frac{s_{2} ^{2} }{n_{2} } ]^{2} }{\frac{(\frac{s_{1} ^{2} }{n_{1} } )^{2}}{n_{1}-1 } + \frac{(\frac{s_{2} ^{2} }{n_{2} } )^{2}}{n_{2}-1 }}

df = \frac{[\frac{3.1^{2} }{38} + \frac{3.8^{2} }{37}]^2}{\frac{(\frac{3.1^{2} }{38})^{2} }{37} + \frac{(\frac{3.8^{2} }{37})^{2} }{36}}

df = 69.4

df = 70, α = 1 - 0.9 = 0.1

t_{\frac{\alpha}{2} } = t_{0.05 }= 1.688

CI = \bar{X_{1} } - \bar{X_{2} } + t_{\frac{\alpha}{2} } \sqrt{\frac{S_{1} ^{2} }{n_{1} +}  \frac{S_{2} ^{2} }{n_{2} } }

CI = (89.2 - 92.4) \pm 1.6883 \sqrt{\frac{3.1^{2} }{38} +\frac{3.8^{2} }{37} }

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