I^27 simplified as much as possible

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

7 0

Answer:

-i. If you are using imaginary numbers.

Step-by-step explanation:

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A random sample of 20 healthy adults was collected and their body temperatures were measured in degrees Fahrenheit. The data can

castortr0y [4]

Answer:

Step-by-step explanation:

Hello!

You need to test the hypothesis that "the average temperature for healthy adults is not equal to 98.6"

The variable of interest is

X: The body temperature of a healthy adult. (Fahrenheit)

And the parameter of interest is the population mean, μ.

The statistic hypotheses are:

H₀: μ = 98.6

H₁: μ ≠ 98.6

α:0.05

Assuming that the variable has a normal distribution you have to use a one-sample t-statistic for this test.

Attached to the answer is the data of body temperature of n=19 healthy adults. Since you didn't copy the raw data for your exercise I'll use this to answer the question.

Using the data the sample mean and standard deviation are:

X[bar]= 98.12

S= 0.69

$t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~~t_{n-1}$

$t_{H_0}= \frac{98.12-98.6}{\frac{0.69}{\sqrt{19} } } = -3.03$

This test is two-tailed, using the critical value approach, you have the rejection region divided into two tails determined by two critical values:

$t_{n-1;\alpha /2}= t_{18;0.025}= -1.965$

$t_{n-1;1-\alpha /2}= t_{18;0.975}= 1.965$

Decision rule:

If $t_{H_0}$ ≤ -1.965 or if $t_{H_0}$ ≥ 1.965, the decision is to reject the null hypothesis.

If -1.965 < $t_{H_0}$ < 1.965, the decision is to not reject the null hypothesis.

The calculated statistic is less than the lower critical value, the decision is to reject the null hypothesis.

Using a significance level of 5%, there is enough evidence to reject the null hypothesis. Then you can conclude that the true mean body temperature for healthy adults is not equal to the traditional 98.6F.

I hope this helps!