Question

A new over-the-counter medicine to treat a sore throat is to be tested for effectiveness. The makers of the medicine take two random samples of 25 individuals showing symptoms of a sore throat. Group 1 receives the new medicine and Group 2 receives a placebo. After a few days on the medicine, each group is interviewed and asked how they would rate their comfort level 1-10 (1 being the most uncomfortable and 10 being no discomfort at all). The results are below. Find the 95% confidence interval in the difference of the mean comfort level. Is there sufficient evidence to conclude that the new medicine is effective? Assume the data is normally distributed with unequal variances.

Average Group 1 = 5.84, SD Group 1 = 2.211334, n1 = 25
Average Group 2 = 3.96, SD Group 2 = 2.35372, n2 = 25

_________ μ1- μ2 _____________

Is there enough evidence to say that the medication is effective?

Answer 1

Answer:

$1.88-2.01\sqrt{\frac{2.211334^2}{25}+\frac{2.35372^2}{25}}=0.582$  

$1.88+2.01\sqrt{\frac{2.211334^2}{25}+\frac{2.35372^2}{25}}=3.178$  

The 95% confidence interval would be given by $0.582 \leq \mu_1 -\mu_2 \leq 3.178$  

For this case we can conclude that we have significant differences between the mean of the two groups and on this case we can conclude that the mean for the group 1(Medicine) is significantly higher than the mean for the group 2(Placebo) at 5% of significance because all the values for the confidence interval are higher than 0. And on this case the medication not effective since the people after the treatment are getting more comfortable

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$\bar X_1 =5.84$ represent the sample mean 1

$\bar X_2 =3.96$ represent the sample mean 2

n1=25 represent the sample 1 size  

n2=25 represent the sample 2 size  

$s_1 =2.211334$ sample standard deviation for sample 1

$s_2 =2.35372$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)  

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =5.84-3.96=1.88$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:  

$df=n_1 +n_2 -1=25+25-2=48$  

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,48)".And we see that $t_{\alpha/2}=2.01$  

Confidence interval

Now we have everything in order to replace into formula (1):  

$1.88-2.01\sqrt{\frac{2.211334^2}{25}+\frac{2.35372^2}{25}}=0.582$  

$1.88+2.01\sqrt{\frac{2.211334^2}{25}+\frac{2.35372^2}{25}}=3.178$  

So on this case the 95% confidence interval would be given by $0.582 \leq \mu_1 -\mu_2 \leq 3.178$  

For this case we can conclude that we have significant differences between the mean of the two groups and on this case we can conclude that the mean for the group 1(Medicine) is significantly higher than the mean for the group 2(Placebo) at 5% of significance because all the values for the confidence interval are higher than 0. And on this case the medication not effective since the people after the treatment are getting more comfortable