Question

A recent study focused on the number of times men and women send a Twitter message in a day. The information is summarized here.Sample Size Sample Mean Population Standard DeviationMen 25 23 5Women 30 28 10At the 0.01 significance level, we ask if there is a difference in the mean number of times men and women send a Twitter message in a day. What is the value of the test statistic for this hypothesis test?

Answer 1

`z=\frac{23-28}{\sqrt{(5^2)/(25)+(10^2)/(30)}}=-2.402`

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and wouldn't be a significant difference in the average for the groups analyzed at the significance level given 1%.

Explanation:

1) Data given and notation

`\bar X_(M)=23` represent the mean for the sample male

`\bar X_(F)=28` represent the mean for the sample female

`\sigma_(M)=5` represent the sample standard deviation for the population male

`\sigma_(F)=10` represent the sample standard deviation for the population female

`n_(M)=25` sample size for the group Stick

`n_(F)=30` sample size for the group Liquid

z would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null hypothesis:
`\mu_(M)=\mu_(F)`

Alternative hypothesis:
`\mu_(M) \\eq \mu_(F)`

Since we have the population deviations given, for this case is better apply a z test to compare means, and the statistic is given by:

`z=\frac{\bar X_(M)-\bar X_(F)}{\sqrt{(\sigma^2_(M))/(n_(M))+(\sigma^2_(F))/(n_(F))}}` (1)

z-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.

In order to calculate the mean and the sample deviation we can use the following formulas:

3) Calculate the statistic

We can replace in formula (1) like this:

`z=\frac{23-28}{\sqrt{(5^2)/(25)+(10^2)/(30)}}=-2.402`

4) Statistical decision

Using the significance level provided
`\alpha=0.01`, we can calculate the p value for this test.

Since is a bilateral test the p value would be:

`p_v =2*P(z<-2.402)=0.016`

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and wouldn't be a significant difference in the average for the groups analyzed at the significance level given 1%.