The branch manager of an outlet (Store 1) of a nationwide chain of pet supply stores wants to study characteristics of her

customers. In particular, she decides to focus on two variables: the

amount of money spent by customers and whether the customers

own only one dog, only one cat, or more than one dog and/or cat.

The results from a sample of 70 customers are as follows:

Amount of money spent: X = +21.34, S = +9.22.
Thirty-seven customers own only a dog.
Twenty-six customers own only a cat.
Seven customers own more than one dog and/or cat.
Construct a 95% confidence interval estimate for the
population mean amount spent in the pet supply store.

Construct a 90% confidence interval estimate for the population proportion of customers who own only a cat.
The branch manager of another outlet (Store 2) wishes to conduct

a similar survey in his store. The manager does not have access to

the information generated by the manager of Store 1. Answer the

following questions:

c. What sample size is needed to have 95% confidence of

estimating the population mean amount spent in this store

to within { +1.50 if the standard deviation is estimated to

be $10?

d. How many customers need to be selected to have 90% confidence of estimating the population proportion of customers who own only a cat to within {0.045?

e. Based on your answers to (c) and (d), how large a sample should the manager take?

1. ( 19.1416, 23.5384,)

2. (0.276348, 0.46651)

3. the sample size = 170.73 approximately 171

4. sample size = 334.07 approximately 334

5. sample of 334 should be taken by manager

Step-by-step explanation:

mean = bar x = 21.34 dollars

size of sample n = 70

standard deviation of sample = 9.22

we use t distribution as the population standard deviation is not known.

95% Confidence interval

1-α = 0.95

α = 0.05

degree of freedom = 70-1 = 69

α/2 = 0.025

using the t distribution tsble,

= 1.9949

confidence interval =

= 21.34 +- (1.9949*1.10200)

= 21.34 + 2.1984, 21.34 - 2.1984

= (23.5384, 19.1416)

the confidence interval of the mean amount spent at the supply store can be written as 19.1416<u<23.5384

2. sixe of those who only have a cat

p = 26/70 = 0.371429

at 90 % confidence interval,

1-α = 0.90

α = 0.10

we use the z table here

z(0.10/2) = Z(0.05)

= 1.645

= 0.371429 +-( 1.645 x 0.0578)

= 0.371429 + 0.095081, 0.371429 - 0.095081

= (0.276348, 0.46651)

3. sd = 10$

margin of erro,r e = 1.50$

α = 0.05

using z table

α/2 = Z0.025

= 1.96

sample size = 1.96² * 10² / 1.50²

= 3.8416 * 100/ 2.25

= 170.73

the sample size is approximately 171

d. we have 0.5 as sample proportion now

margin of error = 0.045

α = 0.10

Zα/2 = 0.05

= 1.645

sample size = 1.645²x0.5(1-0.5) / 0.045²

= 0.676506/0.002025

= 334. 07

sample size = 334

5.  sample of 334 should be taken by manager

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70

the least value out of the data shown is 5, maybe if you post a picture with it i could a little more.

step-by-step explanation: