Suppose your marketing colleague used a known population mean and standard deviation to compute the standard error as 67.5 for samples of a particular size. You don't know the particular sample size but your colleague told you that the sample size is greater than 70. Your boss asks what the standard error would be if you quadruple (4x) the sample size. What is the standard error for the new sample size

The standard error for the new sample size will be of 33.75.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Standard error as 67.5 for samples of a particular size.

We have that , that is, the standard error is inversely proportional to the square root of the sample size, so if you quadruple (4x) the sample size, the standard error will be divided by half. So

67.5/2 = 33.75

The standard error for the new sample size will be of 33.75.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47