In: Statistics and Probability
Due to distance learning and less homework load in Peach Elementary School, parents are complaining that the average screen time is more than 3 hours per day for students in Grade 1 to Grade 3. The principal randomly selected 36 students from Grade 1 - Grade 3 and statistical summary is shown below.
| 
 min  | 
 avg  | 
 max  | 
 std  | 
| 
 3  | 
 5  | 
 8  | 
 2.5  | 
a. Write the hypothesis in symbols or words
b. Check the two conditions for CLT.
c. Calculate the test statistics and the associated degrees of freedom
d. Use the p-value or critical value approach to make your conclusion at 5% significance level.
| 
 one tail (upper)  | 
 0.1  | 
 0.05  | 
 0.025  | 
 0.01  | 
 0.005  | 
| 
 two tails  | 
 0.2  | 
 0.1  | 
 0.05  | 
 0.02  | 
 0.01  | 
| 
 t value, df = 35  | 
 1.31  | 
 1.69  | 
 2.03  | 
 2.44  | 
 2.73  | 
Solution:
Given that
n= 36 sample size
population mean
sample mean
s = 2.5 sample standard deviations
level of significance
a) To test the hypothesis
.
Vs. 
Hypothesis in words
Ho: The average screen time is same 3 hours per day for students in grade1 to grade 3.
Vs
Ha: The average screen time is more than 3 hours per day for students in grade1 to grade 3
b) Condition for CLT
1) The sample observation of grade 1 to grade 3 students are independent to each other.
2) The sample is drawn without replacement of the grede 1 to grade 3 students.
3) sample size of grade 1 to grade 3 students not more than 10% of the population.
c) Test statistic




t = 4.8000007
Test statistic t = 4.8000
The degree of freedom associated with the test statistic is n-1 =36-1=35
The degree of freedom =35
The t critical value at 
 is
from t table
The t critical value =1.69
d) Decision :
t stat > t critical value
4.80 > 1.69
Reject Ho
Conclusion: Reject Ho, there is sufficient evidence to conclude that the average screen time is more than 3 hours per day for students in grade1 to grade 3.