Question

Reference text for questions 1a-1d: Suppose a scratch-off lottery game in NC claims that their cards yield an average prize of $1. To test whether the population mean prize (denote \mu ) is different from 0.25, you obtain a random sample of 81 cards. Then, you find that the mean prize in your sample (denote \overline{x} ) is equal to $0.95, and the sample standard deviation (denote s) is equal to $0.12.

1a. Which (if any) of the following options correctly specifies the null and alternative hypotheses you are testing?

Select one:

a. Null:  = $0.95 ; Alternative:  ≠ $0.95

b. Null:  = $1 ; Alternative:  ≠ $1

c. Null:  = $1 ; Alternative:  ≠ $1

d. Null:  = $0.95 ; Alternative:  ≠ $0.95

e. Null:  =  ; Alternative:  ≠

1b. If the null hypothesis is true, what is the mean of your sampling distribution?

Select one:

a. Mean = $1

b. Mean = $0.95

c. Mean = $1 - $0.95 = $0.05

d. Mean = $0.95 - $1 = -$0.05

1c. Assuming the null hypothesis is true, can you estimate the standard error of your sampling distribution without knowing the population standard deviation?

If so, what does this imply regarding your test statistic and critical values?

Select one:

a. No.

b. Yes. I can use s to estimate my standard error, and a t-distribution to find my test statistic and critical values.

c. Yes. I can use s to estimate my standard error, and a z-distribution to find my test statistic and critical values.

d. Yes. I can use  to estimate my standard error, and a t-distribution to find my test statistic and critical values.

1d. What is the value of your test statistic?

Select one:

a. -3.75

b. -33.75

c. 2.25

d. 3.75

e. Not enough information to calculate.

2. At a significance level of 0.01, you would...

Select one:

a. Reject the null hypothesis.

b. Fail to reject the null hypothesis.

c. Accept the null hypothesis.

d. Accept the alternative hypothesis.

e. Not enough information to conduct the test.

Answer 1

Ho :   µ =   1
Ha :   µ ╪   1
(option b)

.............

mean = 0.95 (option b)

............

c)

b. Yes. I can use s to estimate my standard error, and a t-distribution to find my test statistic and critical values.

d)

sample std dev ,    s =    0.1200                  
Sample Size ,   n =    81                  
Sample Mean,    x̅ =   0.9500                  
                          
degree of freedom=   DF=n-1=   80                  
                          
Standard Error , SE = s/√n =   0.1200   / √    81   =   0.0133      
t-test statistic= (x̅ - µ )/SE = (   0.950   -   1   ) /    0.0133   =   -3.75

option A

..............

critical t value, t* =    ±   2.6387
|t stat | > |critical value | ,

Reject null hypotheis

...........................

Please revert back in case of any doubt.

Please upvote. Thanks in advance.