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lana66690 [7]
3 years ago
10

The mean annual incomes of certified welders are normally distributed with the mean of $50,000 and a standard deviation of $2,00

0. The ship building association wishes to find out whether their welders earn more or less than $50,000 annually. The alternate hypothesis is that the mean is not $50,000. Which of the following is the alternate hypothesis?A. H1: ? ? $50,000.
B. H1: ? < $50,000.
C. H1: ? ? $50,000.
D. H1: ? = $50,000.
Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
6 0

Answer:

Therefore, for the case above; the Alternative hypothesis is;

H1: is not equal to $50,000.

H1: ≠ $50,000.

Step-by-step explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is not different than its mean.

While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

Therefore, for the case above; the Alternative hypothesis is;

H1: is not equal to $50,000.

H1: ≠ $50,000.

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Step-by-step explanation:

Step 1 :

From the histogram we can infer that

0 to 2 texts has been sent by 1 student

2 to 4 texts has been sent by 3 student

4 to 6 texts has been sent by 4 student

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Step 2:

a)

From step 1 , we can see that there are total of 17 students represented by the histogram

Step 3:

b)

This is a skewed histogram because we have all large values of students on the right of the histogram and smaller values on the left of the histogram.

In symmetrical histogram, there would be large values on the center and the smaller values on both sides.

Step 4:

c)

IQR would be better recommendation for this as this is a skewed histogram

Standard deviation would be recommended for a symmetrical histogram

Step 5:

d)

Total number of students is 17

Number of students who text between 8 to 10 messages is 5

Hence the students who text between 8 and 10 messages is 5/17

which is nearly 1/3 rd of the total students.

Hence Sam's observation is correct

Step 6:

e)

Mean amount of the texts = (1 * 3 + 3* 3 + 5*4 +7*4 + 9* 5) /17 = 6.17

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