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.

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.

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step-by-step explanation: