## Researchers measure the body temperature of 52 randomly selected adults. They find a mean temperature of 98.2 degrees with a standard deviat

Question

Researchers measure the body temperature of 52 randomly selected adults. They find a mean temperature of 98.2 degrees with a standard deviation of 0.682 degrees. Which of the following is the correct t-test statistic and p-value for a test of the following hypotheses?
H_o: mu = 98.6 degrees
H_a: mu notequalto 98.6 degrees
The test statistic is negative 1. 039.21.and the p-value is less than 0.000001.
The test statistic is negative 0.46.and the p-value is 2 times P(t_51 > -0.46).
The t-test statistic is negative 0.315.and the p-value is P(t_51 < -0.315).
The t-test statistic is negative 3.33.and the p-value is two times P(t_51 < -3.33).

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4 months 2022-01-31T07:34:19+00:00 1 Answer 0 views 0

The t-test statistic is -3.33.and the p-value is two times P(t_51 < -3.33).

Step-by-step explanation:

Data given and notation

Assuming that the real sample mean is: $$\bar X=98.285$$ represent the sample mean

$$s=0.682$$ represent the sample standard deviation

$$n=52$$ sample size

$$\mu_o =98.6$$ represent the value that we want to test

$$\alpha$$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$$p_v$$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to apply a two tailed test.

What are H0 and Ha for this study?

Null hypothesis: $$\mu = 98.6$$

Alternative hypothesis :$$\mu \neq 98.6$$

Compute the test statistic

The statistic for this case is given by:

$$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$$ (1)

t-test: “Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value”.

Calculate the statistic

We can replace in formula (1) the info given like this:

$$t=\frac{98.2-98.6}{\frac{0.682}{\sqrt{52}}}=-3.298$$

Now we can calculate the degrees of freedom and we got:

$$df = n-1 = 52-1 = 51$$

P value

Since is a two tailed test the p value would be:

$$p_v =2*P(t_{51}<-3.3)=0.0018$$

So the most appropiate conclusion for this case would be:

The t-test statistic is -3.33.and the p-value is two times P(t_51 < -3.33).