Question

the pizza conundrum is the principle that the price of a slice of pizza is always about the same as the subway fare. use the pizza and subway cost data in the table below to determine whether there is a linear correlation between these items. construct a scatterplot, find the value of the linear correlation coefficient r, and find the p - value of r. determine whether there is sufficient evidence to support a claim of linear correlation between these results, does it appear that the subway fare is always about the same as a slice of pizza? use a significance level of α = 0.01
determine the linear correlation coefficient
the linear correlation coefficient is r = 0.990
(round to three decimal places as needed.)
determine the null and alternative hypotheses
h_0: ρ = 0
h_1: ρ ≠ 0
(type integers or decimals. do not round.)
determine the test statistic.
t = □ (round to two decimal places as needed.)

Explanation:

Step 1: Identify Hypotheses

The null hypothesis (\(H_0\)) for a linear correlation test is that there is no linear correlation, so \(H_0:
ho = 0\). The alternative hypothesis (\(H_1\)) for a claim of linear correlation (since we are testing for linear correlation between pizza cost and subway fare) is \(H_1:
ho
eq 0\) (two - tailed test, but if we assume a positive correlation as the scatterplot likely shows a positive trend, but the standard test for linear correlation is two - tailed).

Step 2: Test Statistic Formula

The formula for the test statistic \(t\) for testing the significance of a linear correlation coefficient \(r\) is \(t=\frac{r\sqrt{n - 2}}{\sqrt{1-r^{2}}}\). First, we need to know the sample size \(n\). Let's assume from the pizza and subway fare data (usually, for the pizza connection data, the sample size \(n = 8\) or some common size, but since we know \(r=0.990\), let's proceed. Let's assume \(n\) (number of data points) is, for example, if we consider typical pizza - subway fare data, \(n = 8\).

Step 3: Substitute Values

Given \(r = 0.990\) and \(n=8\) (assuming, if the actual \(n\) is different, the process is similar). First, calculate the numerator: \(r\sqrt{n - 2}=0.990\times\sqrt{8 - 2}=0.990\times\sqrt{6}\approx0.990\times2.4495\approx2.425\)

Then calculate the denominator: \(\sqrt{1 - r^{2}}=\sqrt{1-(0.990)^{2}}=\sqrt{1 - 0.9801}=\sqrt{0.0199}\approx0.1411\)

Now, calculate \(t\): \(t=\frac{2.425}{0.1411}\approx17.19\) (if \(n = 8\)). But let's check with the correct formula. The formula is \(t=\frac{r\sqrt{n - 2}}{\sqrt{1 - r^{2}}}\). Let's assume the sample size \(n\) (number of pairs of data) is, for example, from the pizza - subway fare data, the number of data points is 8 (common in such problems). So \(n-2 = 6\), \(r = 0.990\)

\(t=\frac{0.990\times\sqrt{6}}{\sqrt{1 - 0.990^{2}}}=\frac{0.990\times2.4495}{\sqrt{1 - 0.9801}}=\frac{2.425}{\sqrt{0.0199}}\approx\frac{2.425}{0.141}\approx17.2\) (rounded to two decimal places)

Answer:

If we assume \(n = 8\) (number of data pairs), the test statistic \(t\approx17.20\) (the value may vary slightly depending on the sample size \(n\). If the actual \(n\) is different, recalculate using \(t=\frac{r\sqrt{n - 2}}{\sqrt{1 - r^{2}}}\))