In: Statistics and Probability
National statistics show that the average middle school student spends 8.694 hours per week playing video games. A random sample of 12 middle school students reveals the following times in hours each student spent playing video games.
4.50 9.75 9.50 8.25 5.00 14.50 11.00 4.00 8.25 11.25 11.75 15.50
Has the national average changed? State the hypotheses, check conditions, compute test statistics, standardize and give degrees of freedom. Include a sketch, state the p-value followed by your conclusion.
Here in this scenario our claim is that the average time of playing video games of students is changed from 8.694 hours per week.
Based on given sample of size 12 we have to test this claim Assuming that the level of significance is 0.05 ( the level of significance is not give so we assuming to 0.05).
To test this claim we have to use t distribution one sample t test because here the population standard deviations is unknown.
First we need to compute the sample mean and standard deviations of sample and further the test is performed at 0.05 level of significance as below,
The sample size is n = 12n=12. The provided sample data along with the data required to compute the sample mean \bar XXˉ and sample variance s^2s2 are shown in the table below:
X | X2 | |
4.50 | 20.25 | |
9.75 | 95.0625 | |
9.50 | 90.25 | |
8.25 | 68.0625 | |
5.00 | 25 | |
14.50 | 210.25 | |
11.00 | 121 | |
4.00 | 16 | |
8.25 | 68.0625 | |
11.25 | 126.5625 | |
11.75 | 138.0625 | |
15.50 | 240.25 | |
Sum = | 113.25 | 1218.813 |
The t critical value is calculated using t table or using Excel at 11 degrees of freedom.
Conclusion : since p value is greater than alpha level of significance so we fail to Reject Ho null hypothesis and concluded that there is not enough evidence to support claim that the average is changed.
At 0.05 level of significance our result is not significant.
Thank you.