Statistics
A brochure inviting subscriptions for a new diet program states that the participants are
expected to lose over 22 pounds in five weeks. Suppose that, from the data of the five-week
weight losses of 56 participants, the sample mean and the standard deviation are found to be
23.5 and 10.2 pounds, respectively. Could the statement in the brochure be substantiated on
the basis of these findings? Test with ?=0.05.
A five-year-old census recorded that 20% of the families in a large community lived below the
poverty level. To determine if this percentage has changed, a random sample of 400 families is
studied and 70 are found to be living below the poverty level. Does this finding indicate that the
current percentage of families earning incomes below the poverty level has changed from what
it was five years ago? Test with ?=0.05.
A city health department wishes to determine if the mean bacteria count per unit volume of
water at a lake beach is within the safety level of 200. A researcher collected 10 water samples
of unit volume and found the bacteria counts to be
175, 190, 205, 193, 184, 207, 204, 193, 196, 180
Do the data strongly indicate that there is no cause for concern? Test with ?=0.05.
A limnologist wishes to estimate the mean phosphate content per unit volume of the lake water.
Statistics 2
It is known from studies in previous years that the standard deviation has a fairly stable value of
?=4. How many water samples must the limnologist analyze to be 90% certain that the error of
estimation does not exceed 0.8 milligrams?
Introduction
The tests that are carried out for statistical significance seeks to answer the questions on whether
there is some probability that a relationship exists between two or more variables. Most of the
probabilities are not 100% due to errors such as sampling errors, biasness or other problems with
the reliability of formulas. The hypothesis testing with one sample refers to several tests that
involve the calculations of the z value to obtain the CDF. The following figures have been
obtained by using the formula: z = x – ữ/ϭ/n (Zwillinge & Kokoska, 2010).
1.
1
Mean STDev x z-value CDF
23.5 10.2 22 0.002626 0.501048
The brochure can be substantiated as the chance that the null hypothesis will be rejected is 1-0.5
= 50% while the margin of error is 0.0026 < 0.05 hence it’s acceptable.
Statistics 3
The excel formula used to obtain the Z-value and the CDF are ((P6-R6)/Q6)/56 and
=NORMSDIST (S6) where P6 = 23.5, R6 = 22, Q6 = 10.2 and n = 56 and the formula is Test
statistics is equal to z = x – ữ/ϭ/n
2.
1
Mean STDev x z-value CDF
20 17.5 70 –
0.00714
0.49715
The information can be substantiated as the chance that the null hypothesis will be rejected is 1-
0.5 = 50% while the margin of error is -0.00714 < 0.05 hence it’s acceptable. It can be said that
the current percentages of the families that are earning incomes below the poverty level has
changed from what it was 5 years ago.
The excel formula used to obtain the Z-value and the CDF are ((P6-R6)/Q6)/56 and
=NORMSDIST (S6) where P6 = 20, R6 = 70, Q6 = 17.5 and n = 400 and the formula is Test
statistics is equal to z = x – ữ/ϭ/n
3.
1
Mean STDev x z-value CDF
200 3.54 175 0.0706215 0.507043
200 3.54 190 0.282486 0.611215
200 3.54 205 -0.14124 0.443839
200 3.54 193 0.19774 0.578376
200 3.54 184 0.451977 0.674357
200 3.54 207 -0.19774 0.421624
200 3.54 204 -0.11299 0.455018
200 3.54 193 0.19774 0.578376
200 3.54 196 0.112994 0.544982
Statistics 4
200 3.54 180 0.564972 0.713954
The information can be substantiated as the chance that the null hypothesis will be rejected is 1-
0.71 = 29% while the margin of error is 0.57 > 0.05 hence the null hypothesis will not be
rejected. There is a cause of concern as the data strongly suggests that the margin of error is too
high. It can be said that the mean count of bacteria per unit of water at the lake beach is not
within the safety level of 200 as the margin of error is 0.57 instead of 0.005.
The excel formula used to obtain the Z-value and the CDF are ((P6-R6)/Q6)/10 and
=NORMSDIST (S6) where P6 = 200, R6 = 175, 190, 205, 193, 184,207,204,193,196 and 180
respectively, Q6 = 3.54 and n = 10 and the formula is Test statistics is equal to z = x – ữ/ϭ/n
4.
The formula is equal to n = (z X ϭ/m) ^2
Where z = at 90% confidence = 1.645
Margin of error = 0.8
Standard deviation = ϭ = 4
n = (1.645 X 4/ 0.8) ^2
n = 67.65
Statistics 5
Reference
Zwillinger, D. & Kokoska, S. (2010). CRC Standard Probability and Statistics Tables and
Formulae. CRC Press. p. 49.
