Answer:
There will be sufficient evidence to conclude. A further explanation is provided below.
Step-by-step explanation:
The given values are:
[tex]\sigma=1.8[/tex]
[tex]\alpha=0.05[/tex]
[tex]n=10[/tex]
As we know,
[tex]\bar{x}=\frac{\Sigma x_i}{n}[/tex]
[tex]=\frac{71.5}{10}[/tex]
[tex]=7.15[/tex]
The standard deviation will be:
⇒ [tex]s=\sqrt{\frac{1}{n-1} \Sigma(x_i- \bar{x})^2 }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{1}{10-1} [(6.5-7.15)^2+...(7.7-7.15)^2]}[/tex]
⇒ [tex]=\sqrt{\frac{1}{9} [(6.5-7.15)^2+...(7.7-7.15)^2]}[/tex]
⇒ [tex]=0.477[/tex]
According to the question,
Hypotheses:
[tex]H_o: \sigma=1.8[/tex]
[tex]H_a: \sigma<1.8[/tex]
The test statistic will be:
⇒ [tex]X^2=\frac{(n-1)s^2}{\sigma^2}[/tex]
⇒ [tex]=\frac{(10-1)\times 0.477^2}{1.8^2}[/tex]
⇒ [tex]=\frac{2.0477}{3.24}[/tex]
⇒ [tex]=0.632[/tex]
Thus the above is the correct response.
