Question

A sample of 31 boxes of cereal has a sample standard deviation of 0.81 ounces. Construct a 93% confidence interval to estimate the true standard deviation of the fllins process for the boxes of cereal. a (0433, 1.120) b)(0.658,1.058) c (0.525, 1095) d (0.731,0.731) e) None of the above.

Answer 1

Answer:  Option 'b' is correct.

Step-by-step explanation:

Since we have given that

s = 0.81

n = 31

$s^2=0.6561$

We need to find the 93% confidence interval to estimate the true standard deviation.

degrees of freedom = df = n-1=30

Here, α = 0.93

1-α=0.07

so, $\dfrac{\alpha }{2}=\dfrac{0.07}{2}=0.035$

and $1-\dfrac{\alpha }{2}=1-0.035=0.965$

So, we will use "Chi square distribution"

$\chi^2_{\frac{\alpha }{2},df}=45.455\\\\\chi^2_{1-\frac{\alpha }{2},df}=17.576$

So, the interval would be

$\sqrt{\dfrac{(n-1)s^2}{\chi^2_{\frac{\alpha }{2},df}}}$

Hence, 93% confidence interval would be (0.6581,1.058).

Therefore, Option 'b' is correct.