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3 years ago

To determine whether the means of two populations are equal,

Mathematics

1 answer:

worty [1.4K]3 years ago

4 0

Answer:

The correct option is C. either a t test or an analysis of variance can be performed.

Step-by-step explanation:

Consider the provided information.

The t-test, is used for whether the means of two groups are equal or not. The assumption for the test is that both groups are sampled from normal distributions with equal variances.

Analysis of Variance (ANOVA) is a statistical method evaluating variations between two or more methods. ANOVA is used in a study to analyze the gaps between group methods.

ANOVA is used not for specific differences between means, but for general testing.
The chi-squared test is often used to evaluate whether there was a significant difference in one or more groups between the predicted frequencies and the observed frequencies.

Hence, Either a t test or an analysis of variance can be performed to determine whether the means of two population are equal.

Therefore, the correct option is C. either a t test or an analysis of variance can be performed.

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Henry buys a large boat for the summer, however he cannot pay the full amount of $32,000 at

Anna007 [38]

Answer:

Monthly payments=$418.14

Total amount will be=down payment + 48×$418.14

$14000+$20070.84=$34070.84

Step-by-step explanation:

Loan payment per month=Amount to pay÷discount factor

Mathematically P=A÷D

where D is the discount factor calculated using the formula;

$\frac{(1+i)^n-1}{i(1+i)^n}$

where i=periodic interest rate=annual rate divided by number of payment periods

A is the amount to pay after downpayment

P is the loan monthly payment amount

n=number of periodic payments=payments per year times number of years

⇒In this question you find the discount factor then divide the amount remaining to pay with the discount factor to get monthly payments

Given;

Cost of boat=$32000

Down payment=$14000

Loan to pay=$32000-$14000=$18000

Annual rate=5.5%=i=5.5%÷12=0.458%⇒0.00458

Periodic payments, n=4×12=48

Finding the discount factor D;

$D=\frac{(1+i)^n-1}{i(1+i)^n} \\\\\\D=\frac{(1+0.00458)^{48} -1}{0.00458(1+0.00458)^{48} } \\\\\\D=\frac{1.2455-1}{0.005703} \\\\\\D=\frac{0.2455}{0.005703} =43.05$

To get the amount to pay monthly divide the loan to pay with the discount factor

$=\frac{18000}{43.05} =418.14$

Monthly payments=$418.14

Total amount will be=down payment + 48×$418.14

$14000+$20070.84=$34070.84

8 0

3 years ago

How do you find a solution for an inequality equation?

kozerog [31]

Answer: By adding, subtracting, multiplying or dividing both sides until you are left with the variable on its own. Don't multiply or divide by a variable unless you know it is always positive or always negative.

Step-by-step explanation:

5 0

3 years ago

When we toss a coin, there are two possible outcomes: a head or a tail. Suppose that we toss a coin 100 times. Estimate the appr

marin [14]

Answer:

96.42% probability that the number of tails is between 40 and 60.

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .