Asimple 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 5.25 and the sample standard deviation is 0.55. use a 0.05 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 comma 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:

sample size = n = 60

Sample mean = x bar = 5.25

Sample sd = s = 0.55

Std error of sample = s/sqrt n = 0.071

(One tailed test)

Mean difference = 5.25-5.4 = -0.15

Test statistic t = Mean diff/Se = -2.11

df=59

(since population std dev is not known t test is used)

p value =0.01955

Since p <alpha =0.05 we reject null hypothesis.

The results suggest about the sample group that the sample is from a population with a mean less than 5.4

i am unable to solve this problem

step-by-step explanation:

26 is 33.33% of 78

step-by-step explanation:

26 is what percent of 78?

26 is p% of 78

equation: y = p% * x

solving our equation for p

p% = y/x

p% = 26/78

p = 0.3333

convert decimal to percent:

p% = 0.3333 * 100 = 33.33%