Question

A producer of fine chocolates believes that the sales of two varieties of truffles differ significantly during the holiday season. The first variety is milk chocolate while the second is milk chocolate filled with mint. It is reasonable to assume that truffle sales are normally distributed with unknown but equal population variances. Two independent samples of 18 observations each are collected for the holiday period. A sample mean of 12 million milk chocolate truffles sold with a sample standard deviation of 2.5 million. A sample mean of 13 million truffles filled with mint sold with a sample standard deviation of 2.3 million. Use milk chocolate as population 1 and mint chocolate as population 2. Assuming the population variances are equal, which of the following is the value of the appropriate test statistic?

Answer 1

To find the test statistic for comparing the sales of two varieties of truffles, we use the formula for a two-sample t-test with assumed equal population variances. The test statistic is calculated using the sample means, sample standard deviations, and sample sizes for both types of truffles.

Step-by-step explanation:

The student is asking how to find the value of the test statistic when comparing the means of two independent samples with unknown but equal population variances. Using the provided sample means and standard deviations for the two varieties of milk chocolate truffles—one plain and one filled with mint—we can apply the formula for the test statistic in a two-sample t-test where population variances are assumed to be equal.

The formula is given by:
t = (X₁ - X₂) / S_p * sqrt(1/n₁ + 1/n₂)
where:

• X₁ and X₂ are the sample means for the two populations,
• S_p is the pooled standard deviation,
• n₁ and n₂ are the sample sizes.

First, calculate the pooled standard deviation (S_p) using the formula:
S_p = sqrt(((n₁-1) * S₁² + (n₂-1) * S₂²) / (n₁ + n₂ - 2))
Then, plug the values into the above formula to find the t-statistic.

Answer 2

`\S^2_p =((18-1)(2.5)^2 +(18 -1)(2.3)^2)/(18 +18 -2)=5.77`

`S_p=2.402`

`t=\frac{(12 -13)-(0)}{2.402\sqrt{(1)/(18)+(1)/(18)}}=-1.249`

`p_v =2*P(t_(34)<-1.249) =0.2201`

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that

`\sigma^2_1 =\sigma^2_2 =\sigma^2`

And the statistic is given by this formula:

`t=\frac{(\bar X_1 -\bar X_2)-(\mu_(1)-\mu_2)}{S_p\sqrt{(1)/(n_1)+(1)/(n_2)}}`

Where t follows a t distribution with
`n_1+n_2 -2` degrees of freedom and the pooled variance
`S^2_p` is given by this formula:

`\S^2_p =((n_1-1)S^2_1 +(n_2 -1)S^2_2)/(n_1 +n_2 -2)`

This last one is an unbiased estimator of the common variance
`\sigma^2`

The system of hypothesis on this case are:

Null hypothesis:
`\mu_1 = \mu_2`

Alternative hypothesis:
`\mu_1 \\eq \mu_2`

Or equivalently:

Null hypothesis:
`\mu_1 - \mu_2 = 0`

Alternative hypothesis:
`\mu_1 -\mu_2 \\eq 0`

Our notation on this case :

`n_1 =18` represent the sample size for group 1

`n_2 =18` represent the sample size for group 2

`\bar X_1 =12` represent the sample mean for the group 1

`\bar X_2 =13` represent the sample mean for the group 2

`s_1=2.5` represent the sample standard deviation for group 1

`s_2=2.3` represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

`\S^2_p =((18-1)(2.5)^2 +(18 -1)(2.3)^2)/(18 +18 -2)=5.77`

And the deviation would be just the square root of the variance:

`S_p=2.402`

And now we can calculate the statistic:

`t=\frac{(12 -13)-(0)}{2.402\sqrt{(1)/(18)+(1)/(18)}}=-1.249`

Now we can calculate the degrees of freedom given by:

`df=18+18-2=34`

And now we can calculate the p value using the altenative hypothesis:

`p_v =2*P(t_(34)<-1.249) =0.2201`

If we compare the p value obtained and using the significance level assumed
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.