A new iron supplement is being tested to see if it can increase the mean of ferritin levels. Study participants were divided into two groups. 30 participants in group 1 received the iron supplement; 32 participants in group 2 received a placebo. After one year, group 1 had a mean ferritin level of 43 ng/mL with a standard deviation of 13 ng/mL. Group 2 had a mean ferritin level of 36 ng/mL, with a standard deviation of 11 ng/mL. Can we conclude that the iron supplement increases the population mean of ferritin levels? Use αα=0.005. Note: If t-test, then unequal variances are assumed. Question a) Let: μ1μ1 = the population mean of ferritin levels with the iron supplement μ2μ2 = the population mean of ferritin levels with the placebo (Step 1) State the null and alternative hypotheses by selecting the appropriate symbol, and identify which tailed test: H0:H0: μ1μ1 μ2μ2 H1:H1: μ1μ1 μ2μ2 Which tailed test you need to perform? left-tailed test right-tailed test two-tailed test (Step 2) Find the critical value(s): i) What value is the best criterion to decide which table to use? mean sample size standard deviation ii) Which table (z or t) should be used? tt-table zz-table iii) If you use z-table, then state the closest left cumulative area in the table to find the critical value. If you use t-table, then state 1 tail αα value (in decimal) with the degrees of freedom, separated by a comma. iv) State the critical value(s) : CV = (Step 3) State the rejection region: i) Select from the dropdown menu to shade to the left, to the right, between or left and right of the critical value(s); and then shade the rejection region(s). The arrow can only drag to the critical value(s) that is/are rounded to one decimal place (these values correspond to the tick marks on the horizontal axis). Shade: . Click and drag the arrows to adjust the values. -1.5 ii) Reject H0H0 if test statistic value the critical value(s). (Select >< if outside of two CV or <> if between two CV.) (Step 4) Calculate the test statistic: i) Calculate the standard error which corresponds to the denominator of the test statistic (show 6 decimal places): ii) Calculate the test statistic (rounded to 2 decimal places for z or 3 decimal places for t): test statistic == (Step 5) Decision: Since the test statistic is the critical value(s), we H0H0 at αα= 0.005. (Step 6) Conclusion : Interpret your decision above. There is not enough evidence to conclude that the iron supplement increases the mean of ferritin levels of the participants. There is enough evidence to conclude that the iron supplement increases the population mean of ferritin levels. There is not enough evidence to conclude that the iron supplement increases the population mean of ferritin levels. There is enough evidence to conclude that the iron supplement increases the mean of ferritin levels of the participants. Question b) i) Which type of error could have been made in your conclusion above? ii) Explain the error in that situation. It could be a mistake to reject the null hypothesis when the iron supplement actually decreases or has no effect on the mean of ferritin levels of the participants. It could be a mistake not to reject the null hypothesis when the iron supplement actually decreases or has no effect on the population mean of ferritin levels. It could be a mistake not to reject the null hypothesis when the iron supplement actually increases the population mean of ferritin levels. It could be a mistake to reject the null hypothesis when the iron supplement actually increases the population mean of ferritin levels. It could be a mistake to reject the null hypothesis when the iron supplement actually decreases or has no effect on the population mean of ferritin levels. It could be a mistake not to reject the null hypothesis when the iron supplement actually increses the mean of ferritin levels of the participants.

A new iron supplement is being tested to see if it can increase the mean of ferritin levels. Study participants were divided into two groups. 30 participants in group 1 received the iron supplement; 32 participants in group 2 received a placebo. After one year, group 1 had a mean ferritin level of 43 ng/mL with a standard deviation of 13 ng/mL. Group 2 had a mean ferritin level of 36 ng/mL, with a standard deviation of 11 ng/mL. Can we conclude that the iron supplement increases the population mean of ferritin levels? Use αα=0.005. Note: If t-test, then unequal variances are assumed. Question a) Let: μ1μ1 = the population mean of ferritin levels with the iron supplement μ2μ2 = the population mean of ferritin levels with the placebo (Step 1) State the null and alternative hypotheses by selecting the appropriate symbol, and identify which tailed test: H0:H0: μ1μ1 μ2μ2 H1:H1: μ1μ1 μ2μ2 Which tailed test you need to perform? left-tailed test right-tailed test two-tailed test (Step 2) Find the critical value(s): i) What value is the best criterion to decide which table to use? mean sample size standard deviation ii) Which table (z or t) should be used? tt-table zz-table iii) If you use z-table, then state the closest left cumulative area in the table to find the critical value. If you use t-table, then state 1 tail αα value (in decimal) with the degrees of freedom, separated by a comma. iv) State the critical value(s) : CV = (Step 3) State the rejection region: i) Select from the dropdown menu to shade to the left, to the right, between or left and right of the critical value(s); and then shade the rejection region(s). The arrow can only drag to the critical value(s) that is/are rounded to one decimal place (these values correspond to the tick marks on the horizontal axis). Shade: . Click and drag the arrows to adjust the values. -1.5 ii) Reject H0H0 if test statistic value the critical value(s). (Select >< if outside of two CV or <> if between two CV.) (Step 4) Calculate the test statistic: i) Calculate the standard error which corresponds to the denominator of the test statistic (show 6 decimal places): ii) Calculate the test statistic (rounded to 2 decimal places for z or 3 decimal places for t): test statistic == (Step 5) Decision: Since the test statistic is the critical value(s), we H0H0 at αα= 0.005. (Step 6) Conclusion : Interpret your decision above. There is not enough evidence to conclude that the iron supplement increases the mean of ferritin levels of the participants. There is enough evidence to conclude that the iron supplement increases the population mean of ferritin levels. There is not enough evidence to conclude that the iron supplement increases the population mean of ferritin levels. There is enough evidence to conclude that the iron supplement increases the mean of ferritin levels of the participants. Question b) i) Which type of error could have been made in your conclusion above? ii) Explain the error in that situation. It could be a mistake to reject the null hypothesis when the iron supplement actually decreases or has no effect on the mean of ferritin levels of the participants. It could be a mistake not to reject the null hypothesis when the iron supplement actually decreases or has no effect on the population mean of ferritin levels. It could be a mistake not to reject the null hypothesis when the iron supplement actually increases the population mean of ferritin levels. It could be a mistake to reject the null hypothesis when the iron supplement actually increases the population mean of ferritin levels. It could be a mistake to reject the null hypothesis when the iron supplement actually decreases or has no effect on the population mean of ferritin levels. It could be a mistake not to reject the null hypothesis when the iron supplement actually increses the mean of ferritin levels of the participants.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Question a)
Let: μ1μ1 = the population mean of ferritin levels with the iron supplement
μ2μ2 = the population mean of ferritin levels with the placebo

(Step 1) State the null and alternative hypotheses by selecting the appropriate symbol, and identify which tailed test:

H0:H0: μ1μ1 μ2μ2

H1:H1: μ1μ1 μ2μ2

Which tailed test you need to perform?

left-tailed test
right-tailed test
two-tailed test

(Step 2) Find the critical value(s):

i) What value is the best criterion to decide which table to use?

mean
sample size
standard deviation

ii) Which table (z or t) should be used?

tt-table
zz-table

iii) If you use z-table, then state the closest left cumulative area in the table to find the critical value. If you use t-table, then state 1 tail αα value (in decimal) with the degrees of freedom, separated by a comma.

iv) State the critical value(s) : CV =

(Step 3) State the rejection region:

i) Select from the dropdown menu to shade to the left, to the right, between or left and right of the critical value(s); and then shade the rejection region(s). The arrow can only drag to the critical value(s) that is/are rounded to one decimal place (these values correspond to the tick marks on the horizontal axis).

Shade: . Click and drag the arrows to adjust the values.

-1.5

ii) Reject H0H0 if test statistic value the critical value(s). (Select >< if outside of two CV or <> if between two CV.)

(Step 4) Calculate the test statistic:

i) Calculate the standard error which corresponds to the denominator of the test statistic (show 6 decimal places):

ii) Calculate the test statistic (rounded to 2 decimal places for z or 3 decimal places for t):
test statistic ==

(Step 5) Decision:

Since the test statistic is the critical value(s), we H0H0 at αα= 0.005.

(Step 6) Conclusion : Interpret your decision above.

There is not enough evidence to conclude that the iron supplement increases the mean of ferritin levels of the participants.
There is enough evidence to conclude that the iron supplement increases the population mean of ferritin levels.
There is not enough evidence to conclude that the iron supplement increases the population mean of ferritin levels.
There is enough evidence to conclude that the iron supplement increases the mean of ferritin levels of the participants.

Question b)

i) Which type of error could have been made in your conclusion above?

ii) Explain the error in that situation.

It could be a mistake to reject the null hypothesis when the iron supplement actually decreases or has no effect on the mean of ferritin levels of the participants.
It could be a mistake not to reject the null hypothesis when the iron supplement actually decreases or has no effect on the population mean of ferritin levels.
It could be a mistake not to reject the null hypothesis when the iron supplement actually increases the population mean of ferritin levels.
It could be a mistake to reject the null hypothesis when the iron supplement actually increases the population mean of ferritin levels.
It could be a mistake to reject the null hypothesis when the iron supplement actually decreases or has no effect on the population mean of ferritin levels.
It could be a mistake not to reject the null hypothesis when the iron supplement actually increses the mean of ferritin levels of the participants.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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