Answer:
[tex]\chi^2 =\frac{50-1}{0.2025} 0.3844 =93.015[/tex]
The degrees of freedom are given by:
[tex] df = n-1= 50-1=49[/tex]
And the p value is given by:
[tex]p_v =P(\chi^2 >93.015)=0.00015[/tex]
Since the p value is very low compared to the significance level provided we have enough evidence to conclude that the true deviation is higher than 0.45
Step-by-step explanation:
Information given
[tex]\alpha=0.05[/tex] represent the confidence level
[tex]s^2 =0.62^2 =0.3844 [/tex] represent the sample variance obtained
[tex]\sigma^2_0 =0.45^2 =0.2025[/tex] represent the value that we want to test
System of hypothesis
On this case we want to check if the true deviation is higher than 0.45, so then we can create the following system of hypothesis:
Null Hypothesis: [tex]\sigma^2 \leq 0.2025[/tex]
Alternative hypothesis: [tex]\sigma^2 >0.2025[/tex]
The statistic for this case is given by:
[tex]\chi^2 =\frac{n-1}{\sigma^2_0} s^2[/tex]
Replacing we got
[tex]\chi^2 =\frac{50-1}{0.2025} 0.3844 =93.015[/tex]
The degrees of freedom are given by:
[tex] df = n-1= 50-1=49[/tex]
And the p value is given by:
[tex]p_v =P(\chi^2 >93.015)=0.00015[/tex]
Since the p value is very low compared to the significance level provided we have enough evidence to conclude that the true deviation is higher than 0.45