Question

Cats are visual animals and enjoy watching on-line videos. A new on-line video enhanced for feline viewing is being developed. The mean time that a cat will view the video in one setting is under study. The nine cats in the study viewed the video for 82 seconds on average with a standard deviation 18 seconds. Use this information to answer the questions below.

1. What is the parameter in this situation?
2. What is the point estimate for the parameter in this case?
3. What is the standard error for the point estimator in this case?
4. What is the set of hypotheses that would be used to test that the mean viewing time is 90 seconds against the alternative that the mean viewing time differs from 90 seconds?
5. What is the value of the test statistic to test the null hypothesis that the mean viewing time is 90 seconds?
6. What is the name of the distribution of the test statistic if the mean viewing time really is 90 seconds? This is the set of values possible for the test statistic when the null hypothesis is true.
7. What value must the test statistic be more extreme than in order to reject the null hypothesis at the 1% significance level?
8. What is the value of the p-value in this case? State a range of values, not a specific value. Hint: The p-value is the tail areas associated with the test statistic, that is the area below the negative and the area above the positive.
9. What is the decision about the null hypothesis in this case with a significance level of 1%? Hint: If the p -value is less than the significance level then reject the null hypothesis. The p-value is the chance that you are wrong if you reject the null hypothesis based on these data.
10. What is the conclusion about the alternative hypothesis, based on the decision above about the null hypothesis? Write a sentence that begins, Conclude, the data… .
11. What is the 99% confidence interval to estimate the parameter based on these data?
12. Does the above confidence interval contain the hypothesized parameter value in this case? Write a sentence about what the confidence interval indicates about the parameter value.

Answer 1

Ho :   µ =   90                  
Ha :   µ ╪   90       (Two tail test)          
                          
Level of Significance ,    α =    0.01                  
sample std dev ,    s =    18.0000                  
Sample Size ,   n =    9                  
Sample Mean,    x̅ =   82.0000                  
                          
degree of freedom=   DF=n-1=   8                  
                          
Standard Error , SE = s/√n =   18.0000   / √    9   =   6.0000      
t-test statistic= (x̅ - µ )/SE = (   82.000   -   90   ) /    6.0000   =   -1.3333
                          
critical t value, t* =    ±   3.3554   [Excel formula =t.inv(α/no. of tails,df) ]             
                          
p-Value   =   0.2191   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value>α, Do not reject null hypothesis                       
Conclusion: There is not enough evidence to say that mean time is differ from 90

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margin of error , E=t*SE =   3.3554   *   6.00000   =   20.13
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    82.00   -   20.132324   =   61.867676
Interval Upper Limit = x̅ + E =    82.00   -   20.132324   =   102.132324
99%   confidence interval is (   61.87   < µ <   102.13   )

CI contains the hypothysized parameter value (90)

para meter value is between 61.87 and 102.13

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Please revert back in case of any doubt.

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