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2 years ago

- Given a test statistic of t= 3.018 of a left-tailed test with n= 13, use a 0.01 significance level

Mathematics

1 answer:

alina1380 [7]2 years ago

5 0

Using the t-distribution, it is found that the correct option is:

0.005 < P-value < 0.01, reject the null hypothesis.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if the mean of a given population is equal to 85, that is:

$H_0: \mu = 85$

We have a left-tailed test, hence, at the alternative hypothesis, it is tested if the mean of the population is less than 85, that is:

$H_1: \mu < 85$

<h3>What is the decision?</h3>

The test statistic is of t = -3.018.

Using a t-distribution calculator, with test statistic t = -3.018, 13 - 1 = 12 df and a significance level of 0.01, it is found that the p-value for a left-tailed test is of 0.0051.

The p-value is <u>less than 0.01</u>, hence the null hypothesis is rejected and the correct option is:

0.005 < P-value < 0.01, reject the null hypothesis.

More can be learned about the t-distribution at brainly.com/question/16313918

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Write an inequality to represent the following scenario: Hagrid’s goal at the National Hotdog Eating Contest is to consume 32 ho

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To find the value of a variance add and subtract the variance from the original number.

The original number is 32 hot dogs.

The variance is 5 hot dogs.

32-5 = 27

32 + 5 = 37

This means he needs to eat between 27 and 37 hot dogs.

he inequality with variance is written as: 27 < h < 37

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You decide to put $150 in a savings account to save for a $3,000 down payment on a new car. If the account has an interest rate

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The formula is
A=p (1+r/k)^kt
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R interest rate 0.025
T time?
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12t=Log (3000/150)÷log (1+0.025/12)
T=(log(3,000÷150)÷log(1+0.025÷12))÷12
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3 years ago

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Marcia bought a 50 foot roll she used two 8 and 5/6 how much if left

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Answer:

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Step-by-step explanation:

I suspect two 8 means 8x2, but if it means 2+8, I'll add another answer.

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ORRRRR

2. 2+8=10+.83= 10.83 = 39.17

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3 years ago

The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.10

Salsk061 [2.6K]

For the bell-shaped graph of the normal distribution of weights of Hershey kisses, the area under the curve is 1, the value of the median and mode both is 4.5338 G and the value of variance is 0.0108.

In the given question,

The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.1039 G.

We have to find the answer of many question we solve the question one by one.

From the question;

Mean(μ) = 4.5338 G

Standard Deviation(σ) = 0.1039 G

(a) We have to find for the bell-shaped graph of the normal distribution of weights of Hershey kisses what is the area under the curve.

As we know that when the mean is 0 and a standard deviation is 1 then it is known as normal distribution.

So area under the bell shaped curve will be

$\int\limits^{\infty}_{-\infty} {f(x)} \, dx$ = 1