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Eva8 [605]
3 years ago
10

The temperature at 5:30 A.M. was recorded as 7 degrees below zero on the Fahrenheit scale. Write the temperature using a signed

number​
Mathematics
1 answer:
Svet_ta [14]3 years ago
5 0

Answer:

mhhm yea this is very questioned questioned

Step-by-step explanation:

yea mhm

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your school is setting up a row of 5 tables for a craft fair. each table is 72 inches long. the space between each pair of neigh
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A group of researchers are interested in the possible effects of distracting stimuli during eating, such as an increase or decre
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Using the t-distribution, it is found that since the p-value of the test is of 0.0302, which is <u>less than the standard significance level of 0.05</u>, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

At the null hypothesis, it is <u>tested if the consumption is not different</u>, that is, if the subtraction of the means is 0, hence:

H_0: \mu_1 - \mu_2 = 0

At the alternative hypothesis, it is <u>tested if the consumption is different</u>, that is, if the subtraction of the means is not 0, hence:

H_1: \mu_1 - \mu_2 \neq 0

Two groups of 22 patients, hence, the standard errors are:

s_1 = \frac{45.1}{\sqrt{22}} = 9.6154

s_2 = \frac{26.4}{\sqrt{22}} = 5.6285

The distribution of the differences is has:

\overline{x} = \mu_1 - \mu_2 = 52.1 - 27.1 = 25

s = \sqrt{s_1^2 + s_2^2} = \sqrt{9.6154^2 + 5.6285^2} = 11.14

The test statistic is given by:

t = \frac{\overline{x} - \mu}{s}

In which \mu = 0 is the value tested at the null hypothesis.

Hence:

t = \frac{25 - 0}{11.14}

t = 2.2438

The p-value of the test is found using a <u>two-tailed test</u>, as we are testing if the mean is different of a value, with <u>t = 2.2438</u> and 22 + 22 - 2 = <u>42 df.</u>

  • Using a t-distribution calculator, this p-value is of 0.0302.

Since the p-value of the test is of 0.0302, which is <u>less than the standard significance level of 0.05</u>, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

A similar problem is given at brainly.com/question/25600813

4 0
3 years ago
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