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Scorpion4ik [409]
3 years ago
9

The weight of a USB flash drive is 30 grams and is normally distributed. Periodically, quality control inspectors at Dallas Flas

h Drives randomly select a sample of 17 USB flash drives. If the mean weight of the USB flash drives is too heavy or too light, the machinery is shut down for adjustment; otherwise, the production process continues. The last sample showed a mean and standard deviation of 31.9 and 1.8 grams, respectively. Using α = 0.10, the critical t values are _______.
A. reject the null hypothesis and shut down the process.
B. ​reject the null hypothesis and do not shut down the process.
C. ​fail to reject the null hypothesis and do not shut down the process) do nothing.
D. ​fail to reject the null hypothesis and shut down the process.
Mathematics
1 answer:
Minchanka [31]3 years ago
3 0
I have to think this again. be back later. X=2.22
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Answer:

n = 5 or -5

Step-by-step explanation:

the absolute value is the count away from 0 so n could be 5 which is 5 units from 0 or -5 which is also 5 units away from 0.

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This rectangle is half as wide as it is long what is the perimeter of the rectangle
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length=4 cm width=2 cm

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A psychologist would like to examine the effect of fatigue on mental alertness. An attention test is prepared which requires sub
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Answer:

The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that  fatigue does not significantly increase errors on an attention task at 0.05 significance level

Step-by-step explanation:

We formulate null and alternate hypotheses are

H0 : u1 < u2 against Ha: u1 ≥ u 2

Where u1 is the group tested after they were awake for 24 hours.

The Significance level alpha is chosen to be ∝ = 0.05

The critical region t ≥ t (0.05, 13) = 1.77

Degrees of freedom is calculated df = υ= n1+n2- 2= 5+10-2= 13

Here the difference between the sample means is x`1- x`2= 35-24= 11

The pooled estimate for the common variance σ² is

Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]

= 1/13 [ 120²+360²]

Sp = 105.25

The test statistic is

t = (x`1- x` ) /. Sp √1/n1 + 1/n2

t=   11/ 105.25 √1/5+ 1/10

t=  11/57.65

t= 0.1908

The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that  fatigue does not significantly increase errors on an attention task at 0.05 significance level

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True of false: 0.4&lt; 0.400​
LenaWriter [7]

Answer:

false

Step-by-step explanation:

the are the same jus5 one has more zero afterwards

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-5(5-3v)=-115<br> Helpp me please
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we start by opening the brackets so it will be -25 plus 15 v is equals to 115. negative times a negative is a positive so -5 times -3 we we have gotten positive 15v there I hope you understood so our new equation is -25 + 15v is equals to 115 collect the like terms 15v is equals to 115 + 25 15 v is equals to 140 divide both sides by 15 so v is equals to 9 and 1/3

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