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Artemon [7]
3 years ago
5

or the past 50 days, daily sales of laundry detergent in a large grocery store have been recorded (to the nearest 10). Units Sol

d 30, 40, 50, 60, 70 Number of Times 9, 12, 15, 10, 5 a. Determine the relative frequency for each number of units sold b. Suppose that the following random number were obtained .12 .96 .53 .80 .95 .10 .40 .45 .77 .29 Use these random numbers to simulate the 10 days of sales
Mathematics
1 answer:
bagirrra123 [75]3 years ago
3 0

Answer, step-by-step explanation:

A. With the previous exercise we can deduce that there is the situation of a number of sales in a grocery store, the relative frequency for the number of units sold, is shown below:

units sold. relative frequency. Acumulative frequency. interval of random numbers

30. 0.16. 0.16. 0.00 <0.16

40. 0.24. 0.4. 0.16 <0.4

50. 0.3. 0.7. 0.4 <0.7

60. 0.2. 0.9. 0.7<09

70. 0.1. 1. 0.9<1

B. For the next point, they give us some random numbers and then it is compared with the simulation of 10 days in sales:

random Units

number. sold

0.12. 30

0.96. 70

0.53. 50

0.80. 60

0.95. 70

0.10. 30

0.40. 50

0.45. 50

0.77. 60

0.29. 40

the two lists are compared so that opposite each one is the result of the simulation

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A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A. We are interested in determinin
riadik2000 [5.3K]

Answer:

Option b - not significantly greater than 75%.

Step-by-step explanation:

A random sample of 100 people was taken i.e. n=100

Eighty of the people in the sample favored Candidate i.e. x=80

We have used single sample proportion test,

p=\frac{x}{n}

p=\frac{80}{100}

p=0.8

Now we define hypothesis,

Null hypothesis H_0 : candidate A is significantly greater than 75%.

Alternative hypothesis H_1 : candidate A is not significantly greater than 75%.

Level of significance \alpha=0.05

Applying test statistic Z -proportion,

Z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}

Where, \widehat{p}=80\%=0.80 and p=75%=0.75

Substitute the values,

Z=\frac{0.80-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}

Z=\frac{0.80-0.75}{\sqrt{\frac{0.1875}{100}}}

Z=\frac{0.05}{0.0433}

Z=1.1547

The p-value is

P(Z>1.1547)=1-P(Z

P(Z>1.1547)=1-0.8789

P(Z>1.1547)=0.1241

Now, the p-value is greater than the 0.05.

So we fail to reject the null hypothesis and conclude that the A is not significantly greater than 75%.

Therefore, Option b is correct.

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