Question

For a particular type of restaurant in a large​ city, the food ratings​ (from 0 to 30​ points) had a mean of 21.43 and standard deviation of 3.01. The cost of a dinner​ (in U.S.​dollars) had a mean of ​$79.29 and standard deviation of ​$20.57. The equation that predicts the cost of a dinner using the rating is y hat =0.2+3.7x. The correlation between these two variables is 0.54

a. Predict the cost of a dinner in a restaurant that gets the​ (i) lowest possible rating of​ 0, (ii) highest possible rating of 30.
b. Interpret the slope.
c. Interpret the correlation.
d. Show how the slope can be obtained from the correlation and other information given.

Answer 1

Answer:

a)

i) 0.2

ii) 111.2

b)

Slope=3.7 shows that due to unit increase in ratings the cost of a dinner increases by 3.7 units.

c)

The correlation coefficient 0.54 shows that there is moderate positive relation between ratings and cost of a dinner

d)

slope=r(Sy/Sx).

Slope=0.54(20.57/3.01)

Slope=3.7

Step-by-step explanation:

a.

We are predicting the cost of dinner through rating and the estimated regression equation is

y^=0.2+3.7x.

Where y=cost of dinner and x=ratings

i)

y^=0.2+3.7x

For 0 rating, putting x=0 in above equation

y^ =0.2+3.7(0)

y^=0.2

ii)

y^=0.2+3.7x

For 30 rating, putting x=30 in above equation

y^ =0.2+3.7(30)

y^=0.2 +111

y^=111.2

b.

The regression equation y^=0.2+3.7x shows that slope is 3.7. Slope=3.7 shows that due to unit increase in ratings the cost of a dinner increases by 3.7 units.

c.

The correlation coefficient 0.54 shows that there is moderate positive relation between ratings and cost of a dinner. It means that as ratings increases the cost of a dinner also increases.

d.

We know that

slope=r(Sy/Sx).

We are given that correlation coefficient r is 0.54, standard deviation of cost of dinner Y is 20.57 and the standard deviation of ratings X is 3.01 .

r=0.54, Sy=20.57 and Sx=3.01.

Slope=0.54(20.57/3.01)

Slope=0.54*6.834

Slope=3.69

Thus, slope=3.7.