Answer:
Comparing the p value with a significance level for example we see that
so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the population mean for group 1 is significantly different than the population mean for group 2.
The 95% confidence interval is given by (0.756;7.243), since the interval not contain's the 0 we can reject the null hypothesis that the means are equal.
Step-by-step explanation:
Data given and notation
represent the mean for 1
represent the mean for 2
represent the sample standard deviation for 1
represent the sample standard deviation for 2
sample size for the group 1
sample size for the group 2
Significance level provided
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 population means differs, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:
(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.
With the info given we can replace in formula (1) like this:
P value
We need to find first the degrees of freedom given by:
Since is a two tailed test the p value would be:
Comparing the p value with a significance level for example we see that
so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the population mean for group 1 is significantly different than the population mean for group 2.
Confidence interval
The critical value for this case can be calculated like this. and
. The degrees of freedom are 78 so we can use this code in excel to find the critical value for the interval "=-T.INV(0.025,78)" and we got
The confidence interval for this case would be given by this formula:
And if we replace the values given we have:
The 95% confidence interval is given by (0.756;7.243), since the interval not contain's the 0 we can reject the null hypothesis that the means are equal.