Question

A simple random sample of

60

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

5.24

and the sample standard deviation is

0.51

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

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u > 5.4
Alternative hypothesis: u < 5.4

Note that these hypotheses constitute a one-tailed test.  

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.06584
DF = n - 1

D.F = 59
t = (x - u) / SE

t = - 2.43

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of - 2.43.

Thus the P-value in this analysis is 0.009.

Interpret results. Since the P-value (0.009) is less than the significance level (0.01), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the sample is from a population with a mean less than 5.40.