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melisa1 [442]
1 year ago
8

USE THE FREQUENCY DISTRIBUTION TO DETERMINE THE FOLLOWING:A) THE TOTAL NUMBER OF OBSERVATIONS.B) THE WIDTH OF EACH CLASS.C) THE

MIDPOINT OF THE SECOND CLASS. D) THE MODAL CLASS.E) THE CLASS LIMITS OF THE NEXT CLASS IF AN ADDITIONAL CLASS WERE TO BE ADDEDSEE PHOTO BELOW

Mathematics
1 answer:
lutik1710 [3]1 year ago
4 0

A) The total number is the sum of all the frequencies

2+5+8+12+11+6=44

Answer: 44

B) The width is (upper limit - lower limit) / 2

(35-21) / 2 = 7

(50-36) / 2 = 7

(65-51) / 2 = 7

(80-66) / 2 = 7

(95-81) / 2 = 7

(110-96) / 2 = 7

Answer: the width is 7

C) The midpoint is the lower limit + the width

36+7=43

Answer: the midpoint is 43

D)The modal is the class with more frequency

In this case 66-80 with 12

Answer: class 66-80

E) We use the width formula

lower limit = 111

width = 7

upper = (width * 2) - lower

upper = (7 * 2) + 111

upper = 125

Answer: class 111-125

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Answer:

Null hypothesis:\mu = 13  

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Step-by-step explanation:

Data given and notation  

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\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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State the null and alternative hypotheses to be tested  

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Null hypothesis:\mu = 13  

Alternative hypothesis:\mu \neq 13

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z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

z=\frac{12.2-13}{\frac{2}{\sqrt{30}}}=-2.191  

What do you conclude? Using the p-value approach  

Since is a two tailed test the p value would be:  

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Since the p_v then the correct decision is:

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