Consumer Reports rated 77 cereals on a scale of 0 to 100. The number of grams of

Question

Consumer Reports rated 77 cereals on a scale of 0 to 100. The number of grams of sugar contained in cach serving of the corresponding cereals was also recorded. Using sugar as the explanatory variable and the Consumer Reports rating as the dependent variable, computer output of the data is as follows (the p-values are intentionally left blank) redictorCoef StDev T 5928 | 1948 | 30.43 I-241 | 024 |-10.12! R-Sq . 57.7% R-Sq(adj)-57.1% Sugars s-9.146 Part a: What is the regression equation? Part b: Calculate the 95% confidence interval of the slope of the regression line for all cereals. Part c Use he information provided to test whether there is a significant relationship between the sugar content and the Consumer Report rating at the 5% level

Explanation / Answer

Part c

Here, we have to use t test for population correlation coefficient. The null and alternative hypothesis for this test is given as below:

H0: = 0 versus H0: 0

This is a two tailed test.

We are given

= 5% = 0.05

R-square = 57.7% = 0.577

Correlation coefficient = R = sqrt(0.577) = 0.759605

Sample size = n = 77

t = r*sqrt(n – 2)/sqrt(1 – r^2)

t = 0.759605*sqrt(77 - 2)/sqrt(1 - 0.759605^2)

t = 10.11459

df = n – 2 = 77 – 2 = 75

P-value = 0.00

P-value < = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that there is statistically significant relationship exists between the sugar content and the consumer report rating at 5% level of significance.

Required R codes are given as below:

> Rsquare = 0.577

> R = sqrt(Rsquare)

> n = 77

> t = R*sqrt(n - 2)/sqrt(1 - R^2)

> df = n - 2

> pvalue = 1 - pt(t,df)

> R

[1] 0.7596052

> n

[1] 77

> t

[1] 10.11459

> df

[1] 75

> pvalue

[1] 5.551115e-16

# Reject null

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.