Question

A soft drink manufacturing and exporting company markets a new brand of a soft drink with two experimental flavors. Flavor 1 is sent to thirteen stores; the average sales in the first month is 45 units with a sample standard deviation of 7 units. Flavor 2 is sent to ten stores; the average sales in the first month is 38 units with a sample standard deviation of 6 units. Assume that the populations from which the samples are taken follow normal distribution. State the null and the alternative hypotheses to test whether the data provide sufficient evidence to conclude that the mean sales per month of the soft drink with two flavors are different at a = 0.10.

Answer 1

`t=\frac{(45-38)-0}{\sqrt{(7^2)/(13)+(6^2)/(10)}}}=2.579`

`p_v =2*P(t_(21)>2.579)=0.0175`

Comparing the p value with the significance level
`\alpha=0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two mean sales per month for the 2 flavors is significantly different.

Explanation:

Previous concepts

`\bar X_(1)=45` represent the mean for sample 1

`\bar X_(2)=38` represent the mean for sample 2

`s_(1)=7` represent the sample standard deviation for 1

`s_(f)=6` represent the sample standard deviation for 2

`n_(1)=13` sample size for the group 2

`n_(2)=10` sample size for the group 2

`\alpha=0.1` 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 difference in the population means, the system of hypothesis would be:

Null hypothesis:
`\mu_(1)-\mu_(2)=0`

Alternative hypothesis:
`\mu_(1) - \mu_(2)\\eq 0`

We don't have the population standard deviation's, 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))-\Delta}{\sqrt{(\sigma^2_(1))/(n_(1))+(\sigma^2_(2))/(n_(2))}}` (1)

And the degrees of freedom are given by
`df=n_1 +n_2 -2=13+10-2=21`

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:

`t=\frac{(45-38)-0}{\sqrt{(7^2)/(13)+(6^2)/(10)}}}=2.579`

P value

Since is a bilateral test the p value would be:

`p_v =2*P(t_(21)>2.579)=0.0175`

Comparing the p value with the significance level
`\alpha=0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two mean sales per month for the 2 flavors is significantly different.