Question

1. The is concerned a a water cup when placing an order but fill the cup with a soft drink from the beverage fountain instead of filling the cup with water. Adventure wold manager selected a random sample of 80 customers who bought tickets to the park and found that 23 of these customers bought their tickets at a discounted price. Calculate a 90% confidence interval for the proportion of all customers who, bought their tickets at discount. (please show work).

2. . One of the two cream stores in a hot southern region has two location, one in the city and one in the rural area.. A random sample of 50 customers were selected from the city ice cream store and had a mean buying time of 4.3 minutes with a standard deviation of 3.7 minutes. A random sample of 50 customers from the rural store had a mean buying time of 5.3 minutes with a standard deviation of 3.2 minutes. A 95 percent confidence interval for the difference in mean customer time spent in the store between the two ice cream stores is calculated. What is the standard error of the difference? (please show work)

3. The team leader for a top bakery team has to make a decision on whether to add a new member to the team. She has made an agreement with her team that in order to decide what to do she needs to figure out the average time it takes one team member to bake some cookies, to see if it is longer than 4 minutes. She uses a random sample of baking time for everyone on the team to test the hypotheses ??0: ?? = 4 versus ??1: ?? > 4. Which of the following would be a consequence of a Type I error?

(A) The team leader hires a new person when the baking time is really longer than 4  minutes.

(B) The team leader does not hire a new person additional staff when the baking time is really longer than 4  minutes.

(C) The Team leader hires additional staff when the bake time is not really longer than 4 minutes.

(D) The Team leader does not hire an extra person when the wait time is not really longer than 4 minutes.

4. 50 years ago only 51 percent of British residents with diabetes usually had other illnesses but that amount is significantly higher today.

A Clinician conducted a study to investigate whether the claim was true. The group found that 54 percent of a randomly selected sample of people with diabetes in the United States had other illnesses.

A test of the hypotheses ??0: ?? = 0.51 ??1: ?? > 0.51 resulted in a p-value 0.266.

In these hypotheses, what does ?? represent?

(A) The sample proportion of British individuals that have other illnesses.

(B) The population proportion of British individuals that have other illnesses 50 years ago.

(C) The population proportion of British individuals that have diabetes today. (D) The population mean of British individual that have other illnesses 50 years ago.

Which of the following statements is the best interpretation of the p-value?

(A) The probability that the null hypothesis is true is 0.266.

(B) The probability that the alternative hypothesis is true is 0.266.

(C) If the null hypothesis is true, the probability of observing a sample proportion of at least 0.54 is 0.266.

(D) If the null hypothesis is true, the probability of observing a sample proportion of at most 0.54 is 0.266

Answer 1

1)

Level of Significance,   α =    0.100          
Number of Items of Interest,   x =   23          
Sample Size,   n =    80          
                  
Sample Proportion ,    p̂ = x/n =    0.288          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0506          
margin of error , E = Z*SE =    1.645   *   0.0506   =   0.0832
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.288   -   0.0832   =   0.2043
Interval Upper Limit = p̂ + E =   0.288   +   0.0832   =   0.3707
                  
90%   confidence interval is (   0.204   < p <    0.371   )

--------------------------

2)

Sample #1   ---->     
mean of sample 1,    x̅1=   4.30          
standard deviation of sample 1,   s1 =    3.7          
size of sample 1,    n1=   50          
                  
Sample #2   ---->     
mean of sample 2,    x̅2=   5.300          
standard deviation of sample 2,   s2 =    3.20          
size of sample 2,    n2=   50          
                  
difference in sample means = x̅1-x̅2 =    4.300   -   5.3000   =   -1.0000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.6918         

---------------------------------------

3)

type I error is null hypothesis is rejected when it is true

(C) The Team leader hires additional staff when the bake time is not really longer than 4 minutes.

========================

4)

? represent

(B) The population proportion of British individuals that have other illnesses 50 years ago.

--------------

(C) If the null hypothesis is true, the probability of observing a sample proportion of at least 0.54 is 0.266.