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3. A friend of yours has just started a job as a quality control trainee at a food processing plant. One of the products the

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Answer #1

Here we want to test the hypothesis

H: 0 = 0.3 against H:0 <0.3

If we reject the null hypothesis we can conclude that there is evidence that the regulation is being followed.

It is given that the test statistic is

\chi^2=\frac{(n-1)S^2}{\sigma^2} and for small values of \chi^2 when \sigma=0.3 , we reject the null hypothesis.

From the given data now we want to calculate

S^2=\frac{1}{n-1}\sum(x_i-\bar x)^2

Here

\bar x=\frac{1}{n}\sum x_i=\frac{1}{15}(14.643)=0.9762

S^2=\frac{1}{n-1}\sum(x_i-\bar x)^2=\frac{(1.041 - 0.9762)^2 + (0.662 - 0.9762)^2 +...+ (0.895 - 0.9762)^2}{15-1}\\ \\=\frac{0.4640524}{14}=0.0331466

So the chi square statistic under the null hypothesis becomes

\chi_c^2=\frac{(n-1)S^2}{\sigma_0^2}=\frac{14\times0.0331466}{0.3}\\ \\ =1.547

Suppose the significance level = 0,05

Based on the information provided, at the given significance level = 0,05 , and the the rejection region for this left-tailed test is R= \{\chi^2: \chi^2 < 6.571\} where the number 6.571 is the table value corresponding to = 0,05 from chi square table with 14 degrees of freedom.

Since the calculated value \chi_c^2=1.547<6.571 we reject the null hypothesis and conclude that there is evidence that the regulation is being followed.

____________

Note : If we have

the significance level \alpha=0.01

Based on the information provided, at the given significance level \alpha=0.01 , and the the rejection region for this left-tailed test is R= \{\chi^2: \chi^2 < 4.66\} where the number 4.66 is the table value corresponding to \alpha=0.01 from chi square table with 14 degrees of freedom.

Since the calculated value \chi_c^2=1.547<4.66 we reject the null hypothesis at significance level \alpha=0.01 , same conclusion can be obtained. That is    there is evidence that the regulation is being followed.

___________

If we calculate p-value we get

P(\chi_c^2<\chi_c^2)=P(\chi_c^2<1.547)=0.00002

Since p- value is very close to zero one can conclude that there is evidence that the regulation is being followed.

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