Question

the pizza connection 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 amounts. 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 the two variables. based on 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.
construct a scatterplot. choose the correct graph below
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
h₀: ρ = 0
h₁: ρ ≠ 0
(type integers or decimals. do not round.)
determine the test statistic.
the test statistic is t = □
(round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for test statistic in correlation

The formula for the test statistic \( t \) for testing the significance of the linear correlation coefficient \( r \) is \( t = \frac{r\sqrt{n - 2}}{\sqrt{1 - r^{2}}} \), where \( n \) is the number of pairs of data. But first, we need to know the number of data points. However, since the problem is about testing the linear correlation, and we know \( r = 0.990 \). Let's assume we have \( n \) data points. But maybe from the context, we can proceed. Wait, actually, the null hypothesis \( H_0:
ho = 0 \) and alternative \( H_1:
ho
eq 0 \) (since we are testing for linear correlation, two - tailed test). The test statistic formula is \( t=\frac{r\sqrt{n - 2}}{\sqrt{1 - r^{2}}} \). But maybe in the original problem (since it's about pizza and subway fares, typical data sets for this problem have \( n = 6 \) data points). Let's check with \( n = 6 \).

Step2: Substitute the values into the formula

Given \( r = 0.990 \) and \( n=6 \). Then \( n - 2=6 - 2 = 4 \).

First, calculate the numerator: \( r\sqrt{n - 2}=0.990\times\sqrt{4}=0.990\times2 = 1.98 \)

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

Now, calculate \( t=\frac{1.98}{0.1411}\approx14.03 \) (Wait, but maybe I made a mistake in \( n \). Wait, actually, the standard pizza - subway fare data has \( n = 6 \) pairs. Let's re - check the formula. The formula for the test statistic for correlation is \( t=\frac{r\sqrt{n - 2}}{\sqrt{1 - r^{2}}} \).

If \( r = 0.990 \) and let's assume \( n = 6 \):

\( t=\frac{0.990\times\sqrt{6 - 2}}{\sqrt{1-(0.990)^{2}}}=\frac{0.990\times\sqrt{4}}{\sqrt{1 - 0.9801}}=\frac{0.990\times2}{\sqrt{0.0199}}=\frac{1.98}{0.1411}\approx14.03 \)

But maybe the original problem has \( n = 6 \). Let's confirm.

Alternatively, if we consider the general case, but since the problem is about the "pizza connection" problem, which is a well - known problem with \( n = 6 \) data points (pizza slice cost and subway fare over time, 6 pairs of data).

So, substituting \( r = 0.990 \), \( n = 6 \) into the formula:

\( t=\frac{0.990\times\sqrt{6 - 2}}{\sqrt{1-(0.990)^{2}}}=\frac{0.990\times2}{\sqrt{1 - 0.9801}}=\frac{1.98}{\sqrt{0.0199}}\approx\frac{1.98}{0.141}\approx14.04 \)

Answer:

The test statistic \( t\approx14.04 \) (If \( n = 6 \), the value is approximately 14.04. The exact value depends on the number of data points, but for the typical pizza - subway fare data with \( n = 6 \), the test statistic is approximately 14.04)