Question

A study of the number of business lunches that executives in the insurance and banking industries claim as deductible expenses per month was based on random samples and yielded the following results:

n1 = 40, x = 9.1, s1 = 1.9,
n2 = 50, x = 8.0, s2 = 2.1.

Use the 0.05 level of significance to test the null hypothesis µ1 − µ2 = 0 against the alternative hypothesis µ1 − µ2 ≠ 0.

Answer 1

`t=\frac{9.1-8}{\sqrt{((1.9)^2)/(40)+((2.1)^2)/(50)}}}=2.604`

`p_v =2*P(t_((88))>2.604)=0.0108`

So the p value is a very low value and using any significance level for given
`\alpha=0.05` always
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

Explanation:

Data given and notation

`\bar X_(1)=9.1` represent the mean for the sample 1

`\bar X_(2)=8` represent the mean for the sample 2

`s_(1)=1.9` represent the sample standard deviation for the sample 1

`s_(2)=2.1` represent the sample standard deviation for the sample 2

`n_(1)=40` sample size for the group 1

`n_(2)=50` sample size for the group 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean are different , the system of hypothesis would be:

Null hypothesis:
`\mu_(1) = \mu_(2)`

Alternative hypothesis:
`\mu_(1) \\eq \mu_(2)`

If we analyze the size for the samples both are higher than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

`t=\frac{\bar X_(1)-\bar X_(2)}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}` (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) the results obtained like this:

`t=\frac{9.1-8}{\sqrt{((1.9)^2)/(40)+((2.1)^2)/(50)}}}=2.604`

Statistical decision

The first step is calculate the degrees of freedom, on this case:

`df=n_(1)+n_(2)-2=40+50-2=88`

Since is a bilateral test the p value would be:

`p_v =2*P(t_((88))>2.604)=0.0108`

So the p value is a very low value and using any significance level for given
`\alpha=0.05` always
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%