Question

A recent study focused on the number of times men and women send a WeChat message in a day. The information is summarized next. At the .01 significance level, conclude on whether there is a difference in the mean number of times men and women send a Twitter message in a day.
Sample size sample mean population standard deviation

Men 25 23 5

women 30 28 10

1）Develop the null and alternative hypotheses for this study?

2）Determine a test statistic and the critical value(s).

3) Calculate the value of the test statistic for this hypothesis testing.

4）What is the p-value for this hypothesis test?

5）Based on the p-value, what is your conclusion?

We know the population standard deviation, maybe we should use z test?

Answer 1

Answer:

1)Null hypothesis:$\mu_{men}=\mu_{women}$

Alternative hypothesis:$\mu_{men} \neq \mu_{women}$

2) Two critical values are $z_{\alpha/2}=-2.58$ and $z_{1-\alpha/2}=2.58$

3) $z=-2.40$  

4) $p_v =2*P(z$

5) Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the mean number of times men and women send a Twitter message in a day.

Step-by-step explanation:

Data given and notation

$\bar X_{men}=23$ represent the mean for the sample men

$\bar X_{women}=28$ represent the mean for the sample women

$\sigma_{men}=5$ represent the population standard deviation for the sample men

$\sigma_{women}=10$ represent the population standard deviation for the sample women

$n_{men}=25$ sample size for the group men

$n_{women}=30$ sample size for the group women

t would represent the statistic (variable of interest)

$\alpha=0.01$ significance level provided

1）Develop the null and alternative hypotheses for this study?

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

Null hypothesis:$\mu_{men}=\mu_{women}$

Alternative hypothesis:$\mu_{men} \neq \mu_{women}$

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

$z=\frac{\bar X_{men}-\bar X_{women}}{\sqrt{\frac{\sigma^2_{men}}{n_{men}}+\frac{\sigma^2_{women}}{n_{women}}}}$ (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.

2）Determine the critical value(s).

Based on the significance level$\alpha=0.01$ and $\alpha/2=0.005$ we can find the critical values with the normal standard distribution, we are looking for values that accumulates 0.005 of the area on each tail on the normal distribution.

For this case the two values are $z_{\alpha/2}=-2.58$ and $z_{1-\alpha/2}=2.58$

3) Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$z=\frac{23-28}{\sqrt{\frac{5^2}{25}+\frac{10^2}{30}}}}=-2.40$  

4）What is the p-value for this hypothesis test?

Since is a bilateral test the p value would be:

$p_v =2*P(z$

5）Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the mean number of times men and women send a Twitter message in a day.