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Rufina [12.5K]
3 years ago
10

The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to det

ermine whether or not the mean age of her customers is greater than 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. If she wants to have a level of significance at 0.01, what rejection region should she use?
1. Reject H0 if t < -2.3263.
2. Reject H0 if t > 2.3263.
3. Reject H0 if t < -2.5758.
4. Reject H0 if t > 2.5758.
Mathematics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

Option 2. \text{Reject}~ H_0~ \text{if}~ t > 2.3263

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 30

Sample size, n = 250

Alpha, α = 0.01

First, we design the null and the alternate hypothesis

H_{0}: \mu = 30\\H_A: \mu > 30

We use Right-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Now, t_{critical} \text{ at 0.01 level of significance, 249 degree of freedom } = 2.3263

Decision rule:

For a right ailed t-test,

t_{stat} > t_{critical}, we reject the null hypothesis as it lies in the rejection area.

t_{stat} < t_{critical}, we fail to reject the null hypothesis as it lies in the acceptance area and accept the null hypothesis.

Thus,

Option 2. \text{Reject}~ H_0~ \text{if}~ t > 2.3263

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There were 3232 volunteers to donate blood. Unfortunately, nn of the volunteers did not meet the health requirements, so they co
Degger [83]

Answer:

\frac{470}{32-n}  

Step-by-step explanation:  

We have been given that there were 32 volunteers to donate blood. Unfortunately, n of the volunteers did not meet the health requirements, so they couldn't donate.            

So the number of volunteers that donated blood will be 32-n.

We are also told that the rest of the volunteers donated 470 milliliters each.

To find the units of blood donated by each of the volunteers we will divide total units of donated blood by number of volunteers, who donated the blood (32-n).

\text{Millimeters of blood donated by each volunteer}=\frac{470}{32-n}  

Therefore, each of the volunteers donated \frac{470}{32-n} millimeters of blood.  

5 0
3 years ago
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What #3 and 4 plz help me ^w^
serious [3.7K]
#3
6 x 4 - 4
First multiply 6 x 4
24 - 4
———
5
Subtract the top

20
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7 0
3 years ago
Nd the 52nd term and the term named in the problem <br>40, 140, 240, 340, ...<br>Find a 32​
Galina-37 [17]

Answer:

5140 and 3140

Step-by-step explanation:

Note there is a common difference d between consecutive terms in the sequence, that is

d = 140 - 40 = 240 - 140 = 340 - 240 = 100

This indicates the sequence is arithmetic with n th term

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 40 and d = 100, thus

a_{52} = 40 + (51 × 100) = 40 + 5100 = 5140

a_{32} = 40 + (31 × 100) = 40 + 3100 = 3140

5 0
3 years ago
What is the distance between the two points (5,-2) and (-3,8)
bearhunter [10]
You need to use the distance formula
d =  \sqrt{ {(x - x)}^{2}  +  {(y - y)}^{2} }

\sqrt{ {(5 + 3)}^{2}  +  {( - 2 - 8)}^{2} }
so the distance between points (5,-2) and (-3,8) is
2 \sqrt{41}
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4 0
3 years ago
The article "Occurrence and Distribution of Ammonium in Iowa Groundwater" (K. Schilling, Water Environment Research, 2002:177–18
Papessa [141]

Answer: 95% confidence interval for the difference between the proportions would be (1.31, 1.39).

Step-by-step explanation:

Since we have given that

Number of alluvial wells = 349

Number of quaternary wells = 143

Number of alluvial wells that had concentrations above 0.1 = 182

Number of quaternary wells that had concentrations above 0.1 = 112

Average of alluvial wells = 0.27

Standard deviation = 0.4

Average of quaternary wells = 1.62

Standard deviation =1.70

So, 95% confidence interval gives

alpha = 5% level of significance.

\dfrac{\alpha}{2}=2.5\%\\\\z_{\frac{\alpha}{2}}=1.96

So, 95% confidence interval becomes,

(1.62-0.27)\pm 1.96\sqrt{\dfrac{0.4^2}{349}+\dfrac{1.7^2}{143}}\\\\=1.35\pm 1.96\times 0.020\\\\=(1.35-0.040,1.35+0.040)\\\\=(1.31,1.39)

Hence, 95% confidence interval for the difference between the proportions would be (1.31, 1.39).

6 0
3 years ago
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