Question

The following data are the amounts that a sample of 15 customers spent for lunch​ ($) at a​ fast-food restaurant. Complete parts​ (a) through​ (d) below.

7.48 6.245.8 6.478.34 9.51 7.12 6.85 5.86 4.92 6.49 5.55 7.898.32 9.63

a. Construct a 95% confidence interval estimate for the population mean amount spent for lunch​ ($) at a​ fast-food restaurant.

b. Interpret the interval constructed in​ (a).

c. What assumption must you make about the population distribution in order to construct the confidence interval estimate in​ (a)?

d. Do you think that the assumption needed in order to construct the confidence interval estimate in​ (a) is​ valid? Explain.

Answer 1

a.

= 7.10 , s = 1.412, n = 15

The 95% confidence interval is given by,

( - E, + E )

Where,

$E=t_{c}*rac{s}{sqrt{n}}$

c = 0.95 ,

df = n-1 = 15-1 = 14

( Here we look into t table across df = 14, two tailed probability = 0.05 and find t value )

So,

= 0.78

Hence the required confidence interval is given by,

( 7.10 - 0.78, 7.10 + 0.78 )

( 6.32, 7.88 )

b.Interpretation: We are 95% confident that population mean amount spent for lunch at fast food restaurant lies between 6.32 and 7.88

c.

1. Samples must be drawn randomly .

2. Samples must be independent on each other.

3. Samples must come from normal population.

d. Here we have data from 15 different customers. So randomization condition is satisfied.

Amount spent at restaurant is dependent on what order is placed. The orders placed by one customer is independent of other customers. So, amount spent by these 15 customers at the restaurant is independent of each other.

Restaurant may have large amount of customers out of which we selected 15. So we assume that these large amount of customer follow normal distribution.

So all assumptions are valid.