Question

This term:

This term, out of 85 students who took the beginning-of-term survey, 29 of them love pineapple on pizza. This gives a sample statistic of p-hat = 29/85 = 0.341.

Last two terms:

In the Fall and Spring, out of 60 students that took the beginning-of-term survey, 12 of them love pineapple on pizza. This gives a sample statistic of p-hat = 12/60 = 0.2.

Test a Claim

Conduct a formal hypothesis test for each sample to test Yougov.com’s claim that the population proportion of people in the Western US who like pineapple on pizza is 17%.

For each sample,

1. Write your null and alternative hypotheses in words and symbols. Choose an alpha-level.

2. Check the conditions and assumptions required to approximate the sampling distribution of sample proportions with a normal distribution.

3. What is the center of this normal distribution? What is the standard deviation of this distribution?

4. Compute the p-value of your test statistic, and describe what the p-value means.

5. Find the critical z-values for the rejection region, and then compute your test statistic.

6. Make a decision regarding the null hypothesis.

7. Interpret your decision in the context of this problem.

8. Discuss the consequences of the possible Types of Errors in your conclusion.

Compare Samples

Now use a hypothesis test to compare these samples to each other.

1. Write your null and alternative hypotheses in words and symbols. Choose an alpha-level.

2. Check the conditions and assumptions required to approximate the sampling distribution of differences of sample proportions with a normal distribution.

3. What is the center of this normal distribution? Compute the pooled standard error to approximate the standard deviation.

4. Compute the p-value of your test statistic, and describe what the p-value means.

5. Find the critical z-values for the rejection region, and then compute your test statistic.

6. Make a decision regarding the null hypothesis.

7. Interpret your decision in the context of this problem.

8. Discuss the consequences of the possible Types of Errors in your conclusion.

Answer 1

1)

Ho :   p =    0.17  
H1 :   p ╪   0.17   (Two tail test)
          
Level of Significance,   α =    0.05  
Number of Items of Interest,   x =   29  
Sample Size,   n =    85  
          
Sample Proportion ,    p̂ = x/n =    0.3412  
          
Standard Error ,    SE = √( p(1-p)/n ) =    0.04074  
Z Test Statistic = ( p̂-p)/SE =    (0.3412-0.17)/0.0407=   4.2014  
          
critical z value =    ±    1.960   [excel formula =NORMSINV(α/2)]
          
p-Value   =   0.0000   [excel formula =2*NORMSDIST(z)]
Decision:   p-value<α , reject null hypothesis       
=======================

2)

Ho :   p =    0.17  
H1 :   p ╪   0.17   (Two tail test)
          
Level of Significance,   α =    0.05  
Number of Items of Interest,   x =   12  
Sample Size,   n =    60  
          
Sample Proportion ,    p̂ = x/n =    0.2000  
          
Standard Error ,    SE = √( p(1-p)/n ) =    0.04849  
Z Test Statistic = ( p̂-p)/SE =    (0.2-0.17)/0.0485=   0.6186  
          
critical z value =    ±    1.960   [excel formula =NORMSINV(α/2)]
          
p-Value   =   0.5362   [excel formula =2*NORMSDIST(z)]
Decision:   p value>α ,do not reject null hypothesis       
====================

3)

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 ╪   0          
                  
sample #1   ----->              
first sample size,     n1=   85          
number of successes, sample 1 =     x1=   29          
proportion success of sample 1 , p̂1=   x1/n1=   0.3411765          
                  
sample #2   ----->              
second sample size,     n2 =    60          
number of successes, sample 2 =     x2 =    12          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.200000          
                  
difference in sample proportions, p̂1 - p̂2 =     0.3412   -   0.2000   =   0.1412
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.2827586          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.07593          
Z-statistic = (p̂1 - p̂2)/SE = (   0.141   /   0.0759   ) =   1.8592
                  
z-critical value , Z* =        1.9600   [excel formula =NORMSINV(α/2)]      
p-value =        0.0630   [excel formula =2*NORMSDIST(z)]      
decision :    p-value>α=0.05,Don't reject null hypothesis