Question

The human body maintains its internal temperature through thermoregulation, and a healthy adult has a mean body temperature of 98.6 °F.

The processes of thermoregulation begin to deteriorate in older age, making it difficult to diagnose elderly patients who might have an illness. To understand the body temperature of elderly patients, researchers randomly sampled the body temperatures of elderly patients from a large database of medical records where the body temperatures were normally distributed.

Conduct a two‑tailed, one‑sample ?-test

to determine whether the mean body temperature of the elderly patients is different from that of a healthy adult using the sample data provided (in degrees Fahrenheit). Use a significance level of ?=0.05.

96.6,96.7,96.8,97.4,97.4,97.8,97.8,98.2,98.4,98.9

Answer 1

Solution: For the given problem we construct the null and alternative hypotheses as:

H0: mu = 98.6 vs Ha: mu 98.6

mu = unknown true value of the mean body temperature of the population of healthy adults

The test statistic is T= (xbar-mu0)/(s/sqrt(n)) ; where xbar = sample mean, mu0 = the hypothesized value of the population mean, n = sample size, s = sample standard deviation, sqrt refers to the square root function.

Under H0, T ~ t(n-1)

Here n=10, xbar = 97.6, mu0 = 98.6, s =  0.767391

We reject H0 if |T(observed)| > t(alpha/2,(n-1)), where t(alpha/2,(n-1)) is the upper alpha/2 point of the t - distribution with (n-1) degrees of freedom.   alpha = level of significance.

Here, T(observed) =  -4.120817 and t(alpha/2,(n-1)) = 2.262157(Obtained from the probability table of Student's t distribution)
Hence, |T(observed)| > t(alpha/2,(n-1)).

Hence we reject H0 and conclude at a 5% level of significance on the basis of the given sample that there is enough evidence to support the claim that the the mean body temperature of the elderly patients is different from that of a healthy adult.

[We obtain the answers and critical value using R-software. Code and output are attached below for verification.]