You are constructing a concrete landing pad for a helicopter. You have 4704 cubic feet of concrete

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the

larisa [96]

Answer:

No, at the 0.05 significance level, the number of units produced on the night shift is not larger.

Step-by-step explanation:

We are given that the mean number of units produced by a sample of 54 day-shift workers was 345. The mean number of units produced by a sample of 60 night-shift workers was 351.

Assume the population standard deviation of the number of units produced on the day shift is 21 and 28 on the night shift.

Let $\mu_1$ = population mean number of units produced on the day shift

$\mu_2$ = population mean number of units produced on the night shift

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1\geq\mu_2$ {means that the mean number of units produced on the night shift is same or lesser on the day shift}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1$ {means that the mean number of units produced on the night shift is larger}

The test statistics that will be used here is Two-sample z test statistics as we know about population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }$ ~ N(0,1)

where, $\bar X_1$ = sample mean number of units produced by a sample of 54 day-shift workers = 345

$\bar X_2$ = sample mean number of units produced by a sample of 60 night-shift workers = 351

$\sigma_1$ = population standard deviation of the number of units produced on the day shift = 21

$\sigma_2$ = population standard deviation of the number of units produced on the day shift = 28

$n_1$ = sample of day-shift workers = 54

$n_2$ = sample of night-shift workers = 60

So, test statistics = $\frac{(345-351)-(0)}{\sqrt{\frac{21^{2} }{54}+\frac{28^{2} }{60} } }$

= -1.302

Now at 0.05 significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is more than the critical value of z as -1.302 > -1.6449 so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean number of units produced on the night shift is same or lesser than those produced on the day shift.

4 0

2 years ago

Miguel wrote the predicted and residual values for a data set using the line of best fit y = 1.82x – 4.3. He left out two of the

Dmitry_Shevchenko [17]

Answer:

c

Step-by-step explanation:

i think it was c

5 0

2 years ago

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G(x)= -\dfrac{x^2}{4} + 7g(x)=− 4 x 2 +7g, left parenthesis, x, right parenthesis, equals, minus, start fraction, x, squared,

RUDIKE [14]

Answer:

Step-by-step explanation:

Given the function $G(x)= -\dfrac{x^2}{4} + 7$ , the average rate of change of g(x) over the interval [-2,4], is expressed as shown below;

Rate of change of the function is expressed as g(b)-g(a)/b-a

where a - -2 and b = 4

$G(4)= -\dfrac{4^2}{4} + 7\\G(4)= -\dfrac{16}{4} + 7\\G(4)= -4 + 7\\G(4) = 3\\$

$G(-2) = -\dfrac{(-2)^2}{4} + 7\\G(-2)= -\dfrac{4}{4} + 7\\G(-2)= -1 + 7\\G(-2)= 6$

average rate of change of g(x) over the interval [-2,4] will be;

$g'(x) = \frac{g(4)-g(-2)}{4-(-2)}\\ g'(x) = \frac{3-6}{6}\\\\g'(x) = -3/6\\g'(x) = -1/2$

6 0

2 years ago

Salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2.00m from a waterfall that is 0.55m tall and ju

Law Incorporation [45]

Answer: 6.2 m/s

Explanation:

1) This is a projectile motion (parabolic)

2) Velocity:

i) initial velocity = V₀

ii) Horizontal component:

V₀x = V₀ cos α

The horizontal velocity is constant, so Vx = V₀x

ii) Vertical component:

V₀y = V₀ sin α

The vertical component is linear with acceleration = g ≈ 9.8 m/s²

Vy = V₀y - gt = V₀ sin α - gt

3) Displacement equations

i) Horizontal displacment:

x = V₀ cosα t

ii) Vertical displacement:

y = V₀ sin α t - g t² / 2

y ≈ V₀ sin α t - 4.9 t²

4) Solution

i) x = V₀ cosα t = V₀ cos(32°) t = 2.00 m ← salmon starts 2.00m from a waterfall

⇒ V₀ = 2 / [cos(32°) t ]

ii) y = V₀ sin α t - 4.9 t² = V₀ sin(32°) t - 4.9 t²

iii) Replace V₀ with 2 / [cos(32°) t ]

y = sin(32°) × 2 / [cos(32°) t ] × t - 4.9t² = 2 tan(32°) - 4.9t²

iv) Use jump's height (y = 0.55m) and solve

⇒ 0.55 = 2 tan(32°) - 4.9t²

t² = [2 tan(32°) - 0.55 ] / 4.9 = 0.143 s²

⇒ t = √ (0.143s²) = 0.38 s

v) Use V₀ = 2 / [cos(32°) t ] to find V₀

V₀ = 2 / [cos(32°) (0.38) ] = 6.2 m/s

8 0

2 years ago

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Each day Julian gives his dog, Banjo, 3 liters of water and 500 grams of food. Julian wants to track the total amount of the foo

Masteriza [31]

Answer:

The correct answer would be option C.

Step-by-step explanation:

Since he gives the dog 3 liters of water and 500 grams of food a day we know that on the fifth day the dog will have gotten 15 liters of water and 2,500 grams of food.

4 0

2 years ago

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