A simple random sample of 48 adults is obtained from a normally distributed population, and each person's red blood cell count (in cells per microliter) is measured. The sample mean is 5.23 and the sample standard deviation is 0.56. Use a 0.01 significance level and the given calculator display to test the claim that the sample is from a population with a mean less than 5.4, which is a value often used for the upper limit of the range of normal values. What do the results suggest about the sample group?

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ ≤ 5.4

For the alternative hypothesis,

H1: µ > 5.4

This is a right tailed test

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 48

Degrees of freedom, df = n - 1 = 48 - 1 = 47

t = (x - µ) / (s/√n)

Where

x = sample mean = 5.23

µ = population mean = 5.4

s = samples standard deviation = 0.56

t = (5.23 - 5.4) / (0.56/√48) = 2.1

We would determine the p value using the t test calculator. It becomes

p = 0.021

Since alpha, 0.01 < the p value, 0.021, then we would fail to reject the null hypothesis. Therefore, At a 1% level of significance, there is no significant evidence that the mean red blood cell count of each person is greater than 5.4 cells per microliter.

Since 5.4 is the upper limit, it means that result of the sample group is acceptable.