According to a study conducted by Dell Computers, 65% of men and 70% of women sa

Question

According to a study conducted by Dell Computers, 65% of men and 70% of women say that weight is a very important factor in purchasing a laptop computer. Suppose this survey was conducted using 374 men and 481 women. Test the claim that there is no difference in the proportion of men and women who think weight is a factor in purchasing a laptop? Use a 5% level of significance. The ABC Corporation wants to test for a difference in two methods of training new employees. A sample of 15 new employees is given a three-day seminar (Method A) and a second sample of 12 new employees is given a two-day videocassette course (Method B). Each group is tested at the end of training and the results are: for Method A, the mean is 47.73 and the standard deviation is 4.42, foe Method B, the mean is 56.5 and the standard deviation is 4.27. At a 5% level of significance, can the company assume that there is No difference in the 2 methods of training? What is your conclusion?

Explanation / Answer

1]

Here we want to test that

So null hypothesis is H0 : P1 - P2 = 0

and alternative hypothesis is H1 : P1 - P2 not equal to 0

We can used two sample proportion z test.

level of significance = 0.05

Using minitab.

The command for two sample proportion z test in minitab is

Choose Stat > Basic Statistics > 2 Proportions.

Choose Summarized data.

In First sample, under Events = 243 (we have given 65% men), enter . Under Trials = 374, enter .

In Second sample, under Events = 336 (we have given 70% women), enter . Under Trials = 481, enter .

Click on "Option"

Level of confidence in percentage = c = ( 1- lpha)*100 = = (1 -0.05)*100 = 95.0

so put "Confidence level " = 95.0

Test Difference

Alternative = not equal to

Then click on OK and again click on OK

So we get the following output

Test and CI for Two Proportions

Method

Descriptive Statistics

Estimation for Difference

CI based on normal approximation

Test

The pooled estimate of the proportion (0.677193) is used for the tests.

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.130 > 0.05 so we used 2nd rule.

That is we fail to reject null hypothesis

Conclusion: At 5% level of significance there is no difference in the proportion of men and women who think weight is factor in purchasing laptop.

2]

we have given two sample data sets and here population standard deviations are not known

Two sample mean t test:

First we need to write the given information.

Sample sizes

n1 = 15

n2 = 12

sample means

x1 = 47.73

x2 = 56.5

Sample standard deviations

S1 = 4.42

S2 = 4.27

Let's write the given claim statement.

Claim : There is no difference in the 2 methods of training.

Let's write null and alternative hypothesis from the given claim.

Null hypothesis    H0 : µ1-µ2 = 0

Alternative hypothesis   HA: µ1-µ2 not equal to 0

Using MINITAB

Two-Sample T-Test and CI

Method

Equal variances are not assumed for this analysis.

Descriptive Statistics

Estimation for Difference

Test

Decision rule: 1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.00 < 0.05 so we used first rule.

That is we reject null hypothesis and accept alternative hypothesis.

Conclusion: At 5% level of significance there are sufficient evidence to say that there is significant difference between the 2 methods of training.

p: proportion where Sample 1 = Event p: proportion where Sample 2 = Event Difference: p - p

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