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A company that produces dog food claims that its bags contain 11 kg of food. From
prior experimentation, it is known that the weight of the dog food in these bags is normally
distributed with a standard deviation of 0.6 kg. From a random sample of 25 bags, the mean
weight of the dog food was 10.6 kg. We are concerned that the company is exaggerating how
much dog food they put in their bags. Test at the 5% significance level.

Sagot :

By testing the hypothesis we can conclude that the bag does not contain 11 kg of food.

Given mean of 10.6 kg, population standard deviation of 0.6 and sample size of 25.

We are required to find whether the company is right in saying that their bags contain 11 kg of food.

First we have to make hypothesis.

[tex]H_{0}[/tex]:μ≠11

[tex]H_{1}[/tex]:μ=11

We have to use t statistic because the sample size is less than 30.

t=(X-μ)/s/[tex]\sqrt{n}[/tex]

We will use s/[tex]\sqrt{n}[/tex]=0.6 because we have already given population standard deviation of weights.

t=(11-10.6)/0.6

=0.4/0.6

=0.667

Degree of freedom=n-1

=25-1

=24

T critical at 0.05 with degree of freedom 24=2.0639

T critical at 0.05 with degree of freedom is greater than calculated t so we will accept the null hypothesis.It concludes that the bags donot contain 11 kg of food.

Hence by testing the hypothesis we can conclude that the bag does not contain 11 kg of food.

Learn more about t test at https://brainly.com/question/6589776

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