High school students who take the sat math exam a second time generally score higher than on their first try. past data suggest that the score increase has a standard deviation of about 50 points. how large a sample of high school students would be needed to estimate the mean change in sat score to within 2 points with 92% confidence?

n=1915

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean for the sample  

population mean

represent the population standard deviation  (assumed)

n represent the sample size (variable of interest)  

ME=2 represent the margin of error desired

Confidence =0.92 or 92%

represent the significance level

2) Solution to the problem

The confidence interval for the mean is given by the following formula:

  (1)

The margin of error is given by this formula:

   (a)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

  (b)

The critical value for 92% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.04,0,1)", and we got , replacing into formula (b) we got:

So the answer for this case would be n=1915 rounded up to the nearest integer

don't know sorry bru

step-by-step explanation:

13

step-by-step explanation:

divide the cost of the total of sodas you bought by the cost of one soda, for example:

$24.57 divided by $1.89 is 13 which means you bought 13 sodas