Question

An experiment was conducted to compare the densities (in ounces per cubic inch) of cakes prepared from two different cake mixes. Six cake pans were filled with batter \(\mathrm{A}\), and six were filled with batter B. Expecting a variation in oven temperature, the experimenter placed a pan filled with batter \(\mathrm{A}\) and another with batter \(\mathrm{B}\) side by side at six different locations in the oven. The six paired observations of densities are as follows: $$ \begin{array}{lrrrrrr} \text { Location } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Batter A } & .135 & .102 & .098 & .141 & .131 & .144 \\ \hline \text { Batter B } & .129 & .120 & .112 & .152 & .135 & .163 \end{array} $$ a. Do the data present sufficient evidence to indicate a difference between the average densities of cakes prepared using the two types of batter? b. Construct a \(95 \%\) confidence interval for the difference between the average densities for the two mixes.

Short answer

Explain your answer.

Step-by-step solution

Set up the hypothesis test

We want to test the following hypotheses: \(H_0 : \mu_A = \mu_B\) (There is no difference in the average densities of batters A and B) \(H_1 : \mu_A \neq \mu_B\) (There is a difference in the average densities of batters A and B)

Calculate the differences and their mean

For each location, calculate the difference in densities: \(d_i = x_{A_i} - x_{B_i}\) Differences in densities are: \(0.006, -0.018, -0.014, -0.011, -0.004, -0.019\) Calculate the mean of the differences: \(\bar{d} = \frac{0.006 - 0.018 - 0.014 - 0.011 - 0.004 - 0.019}{6} = -0.01\)

Calculate the standard deviation of the differences

Now, calculate the standard deviation of the differences: \(s_d = \sqrt{\frac{\Sigma (d_i - \bar{d})^2}{n - 1}} = \sqrt{\frac{(0.006 + 0.01)^2 + (-0.018 + 0.01)^2 + ... + (-0.019 + 0.01)^2}{6 - 1}} = 0.0101\)

Calculate the t-statistic

Calculate the t-statistic for the hypothesis test: \(t = \frac{\bar{d} - 0}{\frac{s_d}{\sqrt{n}}} = \frac{-0.01}{\frac{0.0101}{\sqrt{6}}} = -3.16\)

Determine the critical t-value and p-value

The critical t-value for a two-tailed test with 5 degrees of freedom and a significance level of 0.05 (find it with a t-distribution table or calculator): \(t_{critical} = \pm 2.571\) We can also determine the p-value for our t-statistic. Using a t-distribution table or calculator, we find that the p-value is 0.024.

Make a conclusion

Since the calculated t-statistic is less than the critical value (\(-3.16 < -2.571\)), we reject the null hypothesis in favor of the alternative hypothesis. We have sufficient evidence to conclude that there is a difference in the average densities of cakes prepared with batter A and batter B. Now, let's construct the 95% confidence interval for the difference between the average densities.

Calculate the margin of error

To calculate the margin of error, we need to first find the t-value for a 95% confidence level and 5 degrees of freedom: \(t_{0.025} = 2.571\) Now calculate the margin of error using this t-value: \(E = t_{0.025} \times \frac{s_d}{\sqrt{n}} = 2.571 \times \frac{0.0101}{\sqrt{6}} = 0.0107\)

Construct the 95% confidence interval

Now we can construct the 95% confidence interval for the difference between the average densities: \(CI = (\bar{d} - E, \bar{d} + E) = (-0.01 - 0.0107, -0.01 + 0.0107) = (-0.0207, 0.0007)\) The 95% confidence interval for the difference between the average densities of cakes prepared with batter A and batter B is \((-0.0207, 0.0007)\).

Key concepts

Confidence Interval

A confidence interval provides a range of values that is likely to contain the true difference between parameters we're estimating. In this case, it's used to estimate the difference in average densities between two cake batters. By calculating a 95% confidence interval, we're saying that we are 95% confident that the interval we calculate will contain the true difference in average densities. This range helps to understand the precision of our estimate and the reliability of our result.

In the given exercise, after calculating all the necessary statistics, a confidence interval was constructed as (-0.0207, 0.0007).

• The lower bound is -0.0207, suggesting that batter A could potentially produce denser cakes than batter B by about 0.0207 ounces per cubic inch.
• The upper bound is 0.0007, indicating a slight possibility that batter B could yield denser cakes than batter A, but very marginally.

This range includes zero, which is crucial because it signifies that there might be no difference between batters when considering variability. The interval’s proximity to zero indicates the observed difference may not be statistically significant in a broader context.

t-Statistic

The t-statistic is a key element in hypothesis testing, helping us decide whether to reject the null hypothesis. It tells us how many standard deviations our sample mean difference is from the null hypothesis value, which is often zero in these experiments.

To calculate the t-statistic, we use the formula:\[ t = \frac{\bar{d} - 0}{\frac{s_d}{\sqrt{n}}} \]

Here, \(\bar{d}\) represents the mean of the differences, \(s_d\) is the standard deviation of differences, and \(n\) is the number of paired samples. In our exercise, the computed t-statistic is -3.16.

• A negative t-value indicates that the mean difference \(\bar{d}\) is less than zero, implying batter A might result in less dense cakes compared to batter B.
• The magnitude of the t-statistic (-3.16) indicates a substantial deviation from the null hypothesis, leading us to conclude that the difference observed isn’t likely due to random sampling error.

The critical t-value for our test was \(\pm 2.571\) with 5 degrees of freedom, so our calculated t was indeed in the extreme end, allowing us to reject the null hypothesis.

Paired Sample Analysis

Paired sample analysis is a statistical method used to compare two related observations. Unlike independent samples, the data points in paired samples are connected, usually taken from the same subject under different conditions or at two moments in time.

In this exercise, a paired sample analysis was appropriate because each cake mix was subject to the same environmental conditions, reflecting the pairing at each oven location.

• This method involves calculating the differences between each pair of observations. Each location provided a direct comparison for the density between batters A and B.
• By focusing on differences rather than individual values, paired sample analysis controls for potential variability due to extraneous factors, such as oven temperature variation, making the analysis more robust.

The main outcome of such an analysis is a set of difference values, from which we compute the average difference, standard deviation, and perform a hypothesis test. This approach adds sensitivity to the test and enhances the detection of real differences between pairs, evidenced by the conclusion reached in the exercise on batter density.