The weights, in pounds, of the cats in an animal adoption center are normally distributed. If a random sample of cats is taken and the confidence interval is (7.9,12.7), what is the margin of error

Answers

Answer 1
Answer:

Answer: 2.4

Step-by-step explanation:

Given : The weights, in pounds, of the cats in an animal adoption center are normally distributed.

The confidence interval is (7.9,12.7)           (1)

Let \overline{x} be the sample mean of the weights, in pounds, of the cats in an animal adoption center.

We know that the confidence interval for population mean is given by :-

(\overline{x}-E,\overline{x}+E)      (2)

From (1) and (2)

\overline{x}-E=7.9-------(3)\n\n\overline{x}+E=12.7--------------(4)

Subtracting (3) from (4), we get

2E=4.8\n\n\Rightarrow\ E=2.4

Hence, the margin of error is E = 2.4


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Answers

Answer:

Check pdf

Step-by-step explanation:

A company manufacturing computer chips finds that 8% of all chips manufactured are defective. In an effort to decrease the percentage of defective chips, management decides to provide additional training to those employees hired within the last year. After training was implemented, a sample of 450 chips revealed only 27 defects. A hypothesis test is performed to determine if the additional training was effective in lowering the defect rate. Which of the following statement is true about this hypothesis test?a) The additional training significantly increased the defect rate.b) The additional training significantly lowered the defect rate.c) The additional training did affect the defect rate.d) The additional training did not significantly lower the defect rate.e) None of these.

Answers

Answer:

d) The additional training did not significantly lower the defect rate

Step-by-step explanation:

Let proportion of defective chips be = x

Null Hypothesis [H0] : Additional training has no impact on defect rate             x = 8% = 0.08

Alternate Hypothesis [H1] : Additional training has impact on defect rate          x < 8% , x < 0.08  

Observed x proportion (mean) : x' = 27 / 450 = 0.06

z statistic = [ x' - x ]  / √ [ { x ( 1-x ) } / n ]

( 0.06 - 0.08 ) /  √ [ 0.08 (0.92) / 450 ]

= -0.02 / √ 0.0001635  

= -0.02  / 0.01278

z = - 1.56

Since calculated value of z, 1.56 < tabulated value of z at assumed 0.01 significance level, 2.33

Null Hypothesis is accepted, 'training didn't have defect rate reduction impact' is concluded

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Answers

Answer: the answer is 19

Step-by-step explanation:okay so she already have read 119 pages and the total pages is 271 so we substract 271-119=152 so she has 152 pages left so we divide 152 pages divided by 8 because she read 8 pages per hour so 152/8=19

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation: