A group of 80 people who had been diagnosed as prediabetic because of high blood glucose levels volunteered to participate in a study designed to investigate the use of cinnamon to reduce blood glucose to a normal level. Of the 80 people, 40 were randomly assigned to take a cinnamon tablet each day and the other 40 were assigned to take a placebo each day. The people did not know which tablet they were taking. Their blood glucose levels were measured at the end of one month. The results showed that 14 people in the cinnamon group and 10 people in the placebo group had normal blood glucose levels. For people similar to those in the study, do the data provide convincing statistical evidence that the proportion who would be classified as normal after one month of taking cinnamon is greater than the proportion who would be classified as normal after one month of not taking cinnamon?

a. No conclusion can be made about the use of cinnamon because the people in the stuey were volunteers
b. There is convincing statistical evidence at the level of 0.01
c. There is convincing statistical evidence at the level of 0.05 but not at the ver.01
d. There is convincing statistical evidence at the level of 0.10 any reason
e. There is not convincing statistical evidence

The data doesn't provide convincing statistical evidence for considered proportion comparison. We deduce that: Option E: There is not convincing statistical evidence

How to form the hypotheses?

There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.

Null hypothesis is the one which researchers try to disprove.

How to calculate the value of z-test statistic for two sample proportions?

Suppose we have:

[tex]\hat{p}_1[/tex] = first sample's proportion
[tex]n_1[/tex] = first sample's size
[tex]\hat{p}_2[/tex] = second sample's proportion
[tex]n_2[/tex] = first sample's size
[tex]\hat{p}[/tex] = overall proportion

Then, we get:

[tex]Z = \dfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p}_1 )(\dfrac{1}{n_1} + \dfrac{1}{n_2}) }}[/tex]

If the level of significance = [tex]\alpha[/tex]critical value of Z is: [tex]Z_{\alpha/2}[/tex], and if [tex]Z > Z_{\alpha/2}[/tex]null hypothesis, else if [tex]Z_{\alpha/2} > Z[/tex] then we cannot reject it.

For this case, we want to know if the data provides convincing statistical evidence that the proportion who would be classified as normal after one month of taking cinnamon is greater than the proportion who would be classified as normal after one month of not taking cinnamon.

Thus, we get our hypotheses as:

Null hypotheses: They don't provide convincing evidence (and thus, cinnamon had no improvement on the proportion), and thus:

[tex]H_0: \hat{p}_1 \leq \hat{p}_2[/tex]

Alternate hypothesis: Data provides convincing evidence (that the use of cinnamon made increment in considered proportion), and thus:

[tex]H_a: \hat{p}_1 > \hat{p}_2[/tex]

The test is single tailed.

For this case, we have:

[tex]x_1[/tex] = quantity of intended object found in first sample = 14 (of those who take cinnamon tablets each day)
[tex]n_1[/tex] = first sample's size = 40
[tex]\hat{p}_1 = x_1/n_1 = 14/40 =0.35[/tex]
[tex]x_2[/tex] = quantity of intended object found in second sample = 10 (of those who take placebo each day)
[tex]n_2[/tex] = second sample's size = 40
[tex]\hat{p}_2 = x_2/n_2 = 10/40 =0.25[/tex]
[tex]\hat{p}[/tex] overall proportion = [tex](x_1 + x_2)/(n_1 + n_2) = (10 + 14)/80 = 0.3[/tex]

Thus, we get:

[tex]Z = \dfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p}_1 )(\dfrac{1}{n_1} + \dfrac{1}{n_2}) }} =\dfrac{0.35 - 0.25}{\sqrt{0.3(0.7 )(\dfrac{1}{40} + \dfrac{1}{40}) }} \approx 0.976[/tex]

At 0.01 level of significance ([tex]\alpha = 0.01 = 1\%[/tex]), we get the critical value of Z from z-tables as:

[tex]Z_{\alpha/2} = 2.576[/tex]

Since [tex]Z_{\alpha/2} > Z[/tex] , we cannot reject the null hypothesis, and thus, deduce that there is not convincing statistical evidence at the level of 0.01

At 0.05 level of significance ([tex]\alpha = 0.05 = 5\%[/tex]), we get the critical value of Z from z-tables as:

[tex]Z_{\alpha/2} = 1.96[/tex]

Since [tex]Z_{\alpha/2} > Z[/tex] , we cannot reject the null hypothesis, and thus, deduce that there is not convincing statistical evidence at the level of 0.05

At 0.10 level of significance ([tex]\alpha = 0.10 = 10\%[/tex]), we get the critical value of Z from z-tables as:

[tex]Z_{\alpha/2} = 1.645[/tex]

Since [tex]Z_{\alpha/2} > Z[/tex] , we cannot reject the null hypothesis, and thus, deduce that there is not convincing statistical evidence at the level of 0.10

Remember that the conclusions can be made even if they were volunteers since that doesn't make it wrong as volunteers are randomly coming out people from real population.

Thus, we deduce that: Option E: There is not convincing statistical evidence

Learn more about two proportions z-test here:

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