Answer:
The null and alternative hypotheses:
H0: u = 2.8
H1: u < 2.8
This is a left - tailed test.
For test statistics, use the formula:
![Z = (x' - u)/(\sigma/ √(n))](https://img.qammunity.org/2021/formulas/engineering/college/lsknkm34zaaewbqa07lxj4fqeywaxpx581.png)
Where,
Mean, u = 2.8
Sample mean,x' = 2.4
Sample size, n = 61
Standard deviation
= 0.88
Significance level = 0.02
![Z = (2.4 - 0.02)/(0.88/ √(61)) =-3.55](https://img.qammunity.org/2021/formulas/mathematics/high-school/at71c7z47qwyh7nw1bkkw46j276y0jideg.png)
Test statistics, Z = -3.55
Pvalue:
Pvalue of Z = -3.55 using standard normal table,
NORMSTID(-3.55) = 0.00019
Pvalue = 0.0002
Critical value:
Since this is a left tailed test, the critical value at 0.02 level of significance is:
![Z_\alpha = Z_0_._0_2 = -2.05](https://img.qammunity.org/2021/formulas/mathematics/high-school/wq7bkrbuwjopagalir4gz0j55utrq54psg.png)
Decision: If pvalue is less than level of significance reject null hypothesis H0.
Conclusion:
Since pvalue, 0.00019 is less than level of significance, 0.02, reject null hypothesis, H0.
We conclude that the mean life of the cheapest screen is less than 2.8 years