How does the standard deviation help you to determine concentration of the data and whether or not there are potential outliers
1 answer:
You might be interested in
We have similar triangles so
Answer: 35' 3"
Answer:
a. There is evidence that the population mean is above $300.
b. There is no evidence that the population mean is above $300.
c. There is no evidence that the population mean is above $300.
d. The director could ask for cheaper similar books.
Step-by-step explanation:
Let X be the random variable that represents the cost of textbooks. We have observed n = 25 values, = 315.4 and s = 43.20. We suppose that X is normally distributed.
We have the following null and alternative hypothesis
vs
(upper-tail alternative)
We will use the test statistic
and the observed value is
.
If is true, then T has a t distribution with n-1 = 24 degrees of freedom.
a. The rejection region is given by RR = {t | t > } where
is the 90th quantile of the t distribution with 24 df, so, RR = {t | t > 1.3178}. Because the observed value satisty 1.7824 > 1.3178, there is evidence that the population mean is above $300.
b. If s = 75, then the observed value is . The rejection region for a 0.05 level of significance is RR = {t | t >
} where
is the 95th quantile of the t distribution with 24 df, so, RR = {t | t > 1.7108}. Because the observed value does not fall inside the rejection region, there is no evidence that the population mean is above $300.
c. If and s = 43.20, the observed value is
. For RR = {t | t > 1.3178} we have that the observed value does not fall inside RR, therefore, there is no evidence that the population mean is above $300.
d. Because the director of admissions is concerned about the high cost of textbooks, and there is evidence that the population mean of costs is above $300, the director could ask for cheaper similar books.