A two-sample t-test for a difference in means will be conducted to investigate whether the average amount of money spent per customer at Department Store M is different from that at Department Store V. From a random sample of 35 customers at Store M, the average amount spent was $300 with standard deviation $40. From a random sample of 40 customers at Store V, the average amount spent was $290 with standard deviation $35. Assuming a null hypothesis of no difference in population means, what is the test statistic for the appropriate test to investigate whether there is a difference in population means?

The test statistic is t = 1.14.

Step-by-step explanation:

Before testing the hypothesis, we need to understand the central limit theorem and subtraction of normal variables.

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

From a random sample of 35 customers at Store M, the average amount spent was $300 with standard deviation $40.

This means that

From a random sample of 40 customers at Store V, the average amount spent was $290 with standard deviation $35.

This means that

Assuming a null hypothesis of no difference in population means

At the null hypothesis, we test that there is no difference, that is, the subtraction of the means is 0. So

At the alternate hypothesis, we test if there is difference, that is, the subtraction of the means is different of 0. So

The test statistic is:

In which X is the sample mean, is the value tested at the null hypothesis, and s is the standard error.

0 is tested at the null hypothesis:

This means that

From the samples:

What is the test statistic for the appropriate test to investigate whether there is a difference in population means?

The test statistic is t = 1.14.

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.