A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in each can is

Question

A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in each can is 0.12 ounce. The machine is overhauled with new components, and ten cans are filled to determine whether the standard deviation has changed. Assume the fill amounts to be a random sample from a normal population.

Perform a hypothesis test to determine whether the standard deviation differs from 0.12 ounce. Use the level of significance.

in progress 0
Maya 2 months 2021-10-12T04:25:18+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-10-12T04:26:39+00:00

    Answer:

    \chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54

    The degrees of freedom are:

     df=n-1=10-1=9

    Now we can calculate the p value using the alternative hypothesis:

    p_v =2*P(\chi^2 >13.54)=0.279

    Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

    Step-by-step explanation:

    Assuming the following data:”12.14 12.05 12.27 11.89 12.06

    12.14 12.05 12.38 11.92 12.14″

    We can calculate the sample deviation with this formula:

    s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

    n=10 represent the sample size

    \alpha=0.05 represent the confidence level  

    s^2 =0.0217 represent the sample variance

    \sigma^2_0 =0.12^2= 0.0144 represent the value to verify

    Null and alternative hypothesis

    We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:

    Null Hypothesis: \sigma^2 = 0.0144

    Alternative hypothesis: \sigma^2 \neq 0.0144

    The statistic can be calculated like this;

    \chi^2 =\frac{n-1}{\sigma^2_0} s^2

    \chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54

    The degrees of freedom are:

     df=n-1=10-1=9

    Now we can calculate the p value using the alternative hypothesis:

    p_v =2*P(\chi^2 >13.54)=0.279

    Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )