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# Perform a test to establish whether the population mean differ in 1980 and 1998?

In 1980, 350 subjects reported the time they spent every day watching television. The sample mean was 4.1 hours, with standard deviation 3.3. In 1998, 965 subjects reported a mean time spent watching television of 3.7 hours, with a standard deviation 2.0. Perform a test to establish whether the population mean differ in 1980 and 1998?

Again a statistics questions. Any help would be welcome. Thanks.

### 1 Answer

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• Anonymous
1 decade ago
Favorite Answer

Let UA be the population mean of time that people spent everyday watching television in 1998. Then let UB to denote the population mean of time that people spent everyday watching television in 1980.

We wish to set up the following pair of hypothesis to be testd as follows:

We wish to test

H0: UA - UB = 0.

versus

HA:UA- UB ≠ 0.

We are given the following perliminary information: nA=965, xa-bar = 3.7, Sa=2.0, nB= 350, xb-bar = 4.1, Sb = 3.3.

We set to test the above hypothesis at the ∞ = 0.05 level of significance.

Since this test is two-sided, so ∞/2 = 0.05/2 = 0.025.

Z0.025 = 1.96

According to the central limit theorem, since the sample sizes of subjects A and B which are 965 and 350 are both greater than 30, then their sampling distributions would be normally distributed and thus it is appropriate to use the z distributions in this case.

We can only reject H0 in favor of Ha if and only if z>z(∞/2).

The test static = z

= [(SA-bar-SB-bar)- D0]/√[(Sa^2/na^2) + (Sb^2/nb^2)]

= {[(3.7- 4.1)-0]/√[(2)^2/965^2) + (4.1)^2/(350)^2]}

= -33.6

|Z| = |-33.6| = 33.6

Because |z|= 33.6 is much greater than z0.025 = 1.96, we can reject H0 in favor of Ha at the 0.05 level of significance. The p-value for the test is the sum of the area under the standard normal curve to the right and left of z=-33.6 and z=33.6 respectively. This means the p-value = 2* (0.5-0.4990) = 2*0.001 = 0.002. In other words, we have very strong evidence that there is indeed a difference between the population mean in 1980 and 1998.

Hope it helps. I have to go to bed now. Bye..

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