Solve for y: x/y

Which of the following Chi-Square tests can be used to determine if a data set follows a Normal distribution?

shepuryov [24]

Answer:

Answer = d. Chi-Square Goodness of Fit

Step-by-step explanation:

A decision maker may need to understand whether an actual sample distribution matches with a known theoretical probability distribution such as Normal distribution and so on. The Goodness-of-fit Test is a type of Chi-Square test that can be used to determine if a data set follows a Normal distribution and how well it fits the distribution. The Chi-Square test for Goodness-of-fit enables us to determine the extent to which theoretical probability distributions coincide with empirical sample distribution. To apply the test, a particular theoretical distribution is first hypothesized for a given population and then the test is carried out to determine whether or not the sample data could have come from the population of interest with hypothesized theoretical distribution. The observed frequencies or values come from the sample and the expected frequencies or values come from the theoretical hypothesized probability distribution. The Goodness-of-fit now focuses on the differences between the observed values and the expected values. Large differences between the two distributions throw doubt on the assumption that the hypothesized theoretical distribution is correct and small differences between the two distributions may be assumed to be resulting from sampling error.

6 0

3 years ago

Annie buys some greeting cards. Each card costs $1.75. She pays with a twenty dollar bill.

zaharov [31]

Answer:

Step-by-step explanation:

7 0

3 years ago

Factor b2 + b – 6.

Ulleksa [173]

The answer is A,(b + 3)(b - 2)

6 0

3 years ago

The Eco Pulse survey from the marketing communications firm Shelton Group asked individuals to indicate things they do that make

BabaBlast [244]

Answer:

a) 0.54 = 54% probability that a randomly selected person will feel guilty for either wasting food or leaving lights on when not in a room or both.

b) 0.46 = 46% probability that a randomly selected person will not feel guilty for either of these reasons

Step-by-step explanation:

We use Venn's Equations for probabilities.

I am going to say that:

P(A) is the probability that a randomly selected person will feel guilty about wasting food.

P(B) is the probability that a randomly selected person will feel guilty about leaving lights on when not in a room.

0.12 probability that a randomly selected person will feel guilty for both of these reasons.

This means that $P(A \cap B) = 0.12$

0.27 probability that a randomly selected person will feel guilty about leaving lights on when not in a room.

This means that $P(B) = 0.27$

0.39 probability that a randomly selected person will feel guilty about wasting food

This means that $P(A) = 0.39$

a. What is the probability that a randomly selected person will feel guilty for either wasting food or leaving lights on when not in a room or both (to 2 decimals)?

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.39 + 0.27 - 0.12 = 0.54$