The label on a 3-quart container of orange juice states that the orange juice contains an average
of 1 gram of fat or less. answer the following questions for a hypothesis test that
could be used to test the claim on the label.
a. develop the appropriate null and alternative hypotheses.
b. what is the type i error in this situation? what are the consequences of making this error?
c. what is the type ii error in this situation? what are the consequences of making this error?

The null and alternative hypothesis are:

Step-by-step explanation:

Consider the provided information.

Part (A)

The null hypothesis tells the population parameter is equal to the claimed value. If there is no statistical significance in the test then it is called as the null which is denoted by , otherwise it is called as alternative hypothesis which denoted by .

The label on a 3-quart container of orange juice states that the orange juice contains an average  of 1 gram of fat or less.

Hence, the null and alternative hypothesis are:

Part (b)   What is the Type I error in this situation? What are the consequences of making this error?

Type I error is reject the null hypothesis when null hypothesis is true.

If we concluded that mean has increased from 1 gram, while it actually did not.

Therefore, it can be concluded that the label is incorrect, while it is correct.

Part  (c)  What is the Type II error in this situation? What are the consequences of making this error?

Type II error is when the null hypothesis failed to get rejected, when the null hypothesis is false.

If we concluded that the mean has not increased form 1 gram, while it actually did increase.

Therefore, it can be concluded that the label is correct, while it is not.

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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