Question

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?

Answer 1

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.