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it is commonly believed that the mean body temperature of a healthy adu…

Question

it is commonly believed that the mean body temperature of a healthy adult is 98.6°f. you are not entirely convinced. you believe that it is not 98.6°f. you collected data using 50 healthy people and found that they had a mean body temperature of 98.27°f with a standard deviation of 1°f. use a 0.02 significance level to test the claim that the mean body temperature of a healthy adult is 98.6°f.
a. what type of test will be used in this problem? select an answer
b. identify the null and alternative hypotheses?
$h_0$: select an answer?
$h_a$: select an answer?
c. is the original claim located in the null or alternative hypothesis? select an answer
d. calculate your test statistic. write the result below, and be sure to round your final answer to two decimal places.
e. calculate your p - value. write the result below, and be sure to round your final answer to four decimal places.

Explanation:

Step1: Determine test type

Since the population standard - deviation is unknown and we have a sample standard - deviation, we use a one - sample t - test.

Step2: State hypotheses

The null hypothesis $H_0$ is the statement of no effect or the status - quo. The alternative hypothesis $H_a$ is what we are trying to find evidence for. The claim is that the mean body temperature is 98.6°F, so $H_0:\mu = 98.6$ and $H_a:\mu
eq98.6$.

Step3: Locate the claim

The original claim is $\mu = 98.6$, which is in the null hypothesis $H_0$.

Step4: Calculate test statistic

The formula for the one - sample t - test statistic is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$, where $\bar{x} = 98.27$ is the sample mean, $\mu = 98.6$ is the hypothesized population mean, $s = 1$ is the sample standard deviation, and $n = 50$ is the sample size.
\[t=\frac{98.27 - 98.6}{1/\sqrt{50}}\approx\frac{- 0.33}{1/7.07}\approx - 2.33\]

Step5: Calculate p - value

Since this is a two - tailed test with $n-1=49$ degrees of freedom and $t=-2.33$, using a t - distribution table or technology, the p - value is $2P(T > |-2.33|)$ with $df = 49$. Using a calculator or software, the p - value is approximately 0.0242.

Answer:

a. One - sample t - test
b. $H_0:\mu = 98.6$, $H_a:\mu
eq98.6$
c. Null hypothesis
d. - 2.33
e. 0.0242