A recent survey conducted by Wall Street Journal indicates that the average reading time of 100 its readers/participants was 45 minutes and the standard deviation was 20 minutes. The journal is now planning another survey study. If the acceptable error is 3 minutes and a confidence level of 99% is desired, what would be the minimum sample size needed for the upcoming survey?

The minimum sample size needed for the upcoming survey

n = 289

Step-by-step explanation:

Explanation:-

step:- (i)

Given A recent survey conducted by Wall Street Journal indicates that the average reading time of 100 its readers/participants was 45 minutes and the standard deviation was 20 minutes

sample size n= 100

mean =45 min and standard deviation = 20 minutes

If the acceptable error is 3 minutes and a confidence level of 99% is desired

Given margin of error is '3' min

Step(ii):-

Margin error

We know that margin of error  =

Now we determine the sample size

The 99% of level of significance 'Z' value = 2.57

n = (17)^2

n = 289

Conclusion:-

The minimum sample size needed for the upcoming survey

n = 289

Verification:-

Margin error =

Given data of margin error also =3

so both are equal

There fore the minimum sample size needed for the upcoming survey

n = 289

the answer is d. (1.2, -4.7) on edgn

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answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.