In Exercises 5–8, use a significance level 0.05 and refer to theaccompanying displays.Cereal Killers The amounts of sugar (grams of sugar per gram of cereal) and calories (per gram of cereal) were recorded for a sample of 16 different cereals. TI-83>84 Plus calculator results are shown here. Is there sufficient evidence to support the claim that there is a linear correlation between sugar and calories in a gram of cereal? Explain.

Two variables are being studied:the amount of sugar (grams of sugar per gram of cereal), and calories (per gram of cereal).

The sample size of cereals is 16(n).

The output gives the correlation coefficient as 0.7654038409.

Let\(\rho \)be the true correlation coefficient measure for the amount of sugar and calories.

For testing the claim, form the hypotheses as shown:

\(\begin{array}{l}{{\rm{H}}_{\rm{o}}}:\rho = 0\\{{\rm{{\rm H}}}_{\rm{a}}}:\rho \ne 0\end{array}\)

The samples number of cereals is16(n).

The test statistic is computedbelow:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.7654038409}}{{\sqrt {\frac{{1 - {{0.7654038409}^2}}}{{16 - 2}}} }}\\ &= 4.4501\end{aligned}\)

Thus, the test statistic is 4.450.

The degree of freedom is computedbelow:

\(\begin{aligned} df &= n - 2\\ &= 16 - 2\\ &= 14\end{aligned}\)

The p-value is computed using thet-distribution table.

\(\begin{aligned} p{\rm{ - value}} &= P\left( {T > t} \right)\\ &= 2P\left( {T > 4.4501} \right)\\ &= 2\left( {1 - P\left( {T < 4.4501} \right)} \right)\\ &= 0.0005\end{aligned}\)

Thus, the p-value is 0.0005.

Since the p-value is lesser than 0.05, the null hypothesis is rejected.

Thus, there is enough evidence to support theexistence of a linear association between the amount of sugar (grams of sugarper gram of cereal)and calories (per gram of cereal).