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Debora [2.8K]
3 years ago
11

Four brands of lightbulbs are being considered for use in the final assembly area of the Ford F-150 truck plant in Dearborn, Mic

higan. The director of purchasing asked for samples of 200 from each manufacturer. The numbers of acceptable and unacceptable bulbs from each manufacturer are shown below. At the 0.10 significance level, is there a difference in the quality of the bulbs?
Manufacturer
A B C D
Unacceptable 29 17 9 22
Acceptable 171 183 191 178
Total 200 200 200 200
H0: There is no relationship between quality and manufacturer.
H1: There is a relationship.
1) State the decision rule using 0.10 significance level. (Round your answer to 3 decimal places.)
Reject H0 if chi-square >
2) Compute the value of chi-square. (Round your answer to 3 decimal places.)
Chi-square value
Mathematics
1 answer:
nlexa [21]3 years ago
6 0

Answer:

Decison region :

Reject H0 : if χ² > 6.251

12.229

Step-by-step explanation:

Given :

Manufacturer A B C D

Unacceptable 29 17 9 22

Acceptable 171 183 191 178

Total 200 200 200 200

H0: There is no relationship between quality and manufacturer.

H1: There is a relationship.

Testing using the goodness of fit :

Chisquare = (observed - Expected)² / Expected

Expected Values:

19.25 19.25 19.25 19.25

180.75 180.75 180.75 180.75

Chi-Squared Values:

4.93831 0.262987 5.45779 0.392857

0.525934 0.0280083 0.581259 0.0418396

χ² = 4.93831 + 0.262987 + 5.45779 + 0.392857

+ 0.525934 + 0.0280083 + 0.581259 + 0.0418396 = 12.229

Degree of freedom, df = (4-1)(2-1) = 3*1= 3

The critical value,

χ² at 0.10, 3 = 6.251

Decison region :

Reject H0 : if χ² > 6.251

Reject H0 : 12.229 > 6.251

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nexus9112 [7]

Answer:

46 books

Step-by-step explanation:

276/6=46 books per shelf

8 0
3 years ago
A random sample of 50 cars in the drive-thru of a popular fast food restaurant revealed an average bill of $18.21 per car. The p
mars1129 [50]

Answer:

a) \bar{x} = 18.21

b) 1.64

c) (16.57,19.85)

d) The sample size must be 135 or greater if the error must not exceed $1.00.

Step-by-step explanation:

We are given the following information in the question:

Sample size, n = 50

Sample mean =   $18.21

population standard deviation = $5.92

a)  Point estimate for the population mean cost

\bar{x} = 18.21

b) Margin of error =

z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

Margin of error =  1.96\displaystyle\frac{5.92}{\sqrt{50}} = 1.64

c) 95% Confidence interval

\mu \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

18.21 \pm 1.64) = (16.57,19.85)

d) Marginal error less than $1.00

1.96\displaystyle\frac{\sigma}{\sqrt{n}} \leq 1\\\\\sqrt{n} \geq 1.96\times 5.92\\n \geq (11.6)^2\\n \geq 134.56 \approx 135

Thus, the sample size must be 135 or greater if the error must not exceed $1.00.

5 0
3 years ago
1. Solve the following simultaneous equations using the matrix method.
dmitriy555 [2]

(i) Use the formula for the determinant of a 2×2 matrix.

\begin{vmatrix}a&b\\c&d\end{vmatrix} = ad-bc

\implies \det(A) = \begin{vmatrix}4 & -3 \\ 2 & 5\end{vmatrix} = 4\times5 - (-3)\times2 = \boxed{26}

(ii) The adjugate matrix is the transpose of the cofactor matrix of A. (These days, the "adjoint" of a matrix X is more commonly used to refer to the conjugate transpose of X, which is not the same.)

The cofactor of the (i, j)-th entry of A is the determinant of the matrix you get after deleting the i-th row and j-th column of A, multiplied by (-1)^{i+j}. If C is the cofactor matrix of A, then

C = \begin{pmatrix}5&-2\\3&4\end{pmatrix}

Then the adjugate of A is the transpose of C,

\mathrm{adj}(A) = C^\top = \boxed{\begin{pmatrix}5&3\\-2&4\end{pmatrix}}

(iii) The inverse of A is equal to 1/det(A) times the adjugate:

A^{-1} = \dfrac1{\det(A)} \mathrm{adj}(A) = \boxed{\begin{pmatrix}\frac5{26}&\frac3{26}\\\\-\frac1{13}&\frac2{13}\end{pmatrix}}

(iv) The system of equations translates to the matrix equation

A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}6\\16\end{pmatrix}

Multiplying both sides on the left by the inverse of A gives

A^{-1}\left(A\begin{pmatrix}x\\y\end{pmatrix}\right)=A^{-1} \begin{pmatrix}6\\16\end{pmatrix}

\left(A^{-1}A\right)\begin{pmatrix}x\\y\end{pmatrix}=A^{-1} \begin{pmatrix}6\\16\end{pmatrix}

\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\frac5{26}&\frac3{26}\\\\-\frac1{13}&\frac2{13}\end{pmatrix} \begin{pmatrix}6\\16\end{pmatrix}

\begin{pmatrix}x\\y\end{pmatrix}=\boxed{\begin{pmatrix}3\\2\end{pmatrix}}

4 0
2 years ago
Raj and Aditi like to play chess against each other. Aditi has won 12 of the 20 games so far. They decide to play more games. Ho
Over [174]
12 + x = 0.80(20+x)
12 + x = 16 + 0.80x
12 = 16 -0.2x
-4 = -0.2x

x = -4/-0.2 = 20
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4 0
3 years ago
Read 2 more answers
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NNADVOKAT [17]

Answer:

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Segment Adjacent to the leg: 36

x=60

Step-by-step explanation:

I submitted and it was correct

6 0
3 years ago
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