Question
Jan Villaroel
Topic: Statistics Posted 1 year ago
When the tissue around teeth becomes inflamed, the tissue bleeds more easily and this is an indication of gum disease. The severity of bleeding

When the tissue around teeth becomes inflamed, the tissue bleeds more easily and this is an indication of
gum disease. The severity of bleeding can be measured by a bleeding score that is determined by the
percentage of teeth for which bleeding occurs when the tissue surrounding the tooth is probed. In terms of
dental health, lower bleeding scores are better than higher bleeding scores. The authors of a paper
wondered if the nutrients found in kiwis might reduce bleeding. To investigate, 25 adult participants agreed
to consume 2 kiwis a day for a period of 2 months. A bleeding score was recorded for each participant at
the beginning of the study period and again after 2 months of kiwi consumption. Because this resulted in
paired samples, differences in bleeding scores (before − after) were analyzed. It was reported that the
sample mean difference was xd = 6.68 and the standard deviation of the differences was sd = 11.8
Assuming that the 25 study participants can be considered as representative of the population of adults, is
there convincing evidence that the mean bleeding score is lower after 2 months of kiwi consumption? Test
the relevant hypotheses using a significance level of alpha = 0.01. (Use 𝜇1 for before kiwi consumption, 𝜇2 for after kiwi consumption, and 𝜇d = 𝜇1 − 𝜇2 .)

a) State the appropriate null and alternative hypotheses (in minutes).

b) Find the test statistic.

c) Find p-value


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Koala Tutor
Dec 14, 2023
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a) The appropriate null and alternative hypotheses are as follows:

Null hypothesis (H0): The mean bleeding score before kiwi consumption is equal to the mean bleeding score after kiwi consumption (𝜇1 = 𝜇2).

Alternative hypothesis (Ha): The mean bleeding score before kiwi consumption is lower than the mean bleeding score after kiwi consumption (𝜇1 < 𝜇2).

b) The test statistic can be calculated using the formula:

t = (xd - Ξd) / (sd / sqrt(n))

where
- xd is the sample mean difference (6.68),
- Ξd is the population mean difference (0, based on the null hypothesis),
- sd is the standard deviation of the differences (11.8), and
- n is the number of participants (25).

Plugging in the values, we get:

t = (6.68 - 0) / (11.8 / sqrt(25))
t = 6.68 / (11.8 / 5)
t ≈ 6.68 / 2.366

Therefore, the test statistic is approximately 2.82.

c) To find the p-value, we need to compare the test statistic to the appropriate t-distribution with n-1 degrees of freedom. In this case, the degrees of freedom is 25-1 = 24.

At a significance level of alpha = 0.01, we need to find the p-value for a one-tailed test in the left tail of the t-distribution.

Using a t-distribution table or calculator, we find that the p-value for t = 2.82 (with 24 degrees of freedom) is approximately 0.005.

Since the p-value (0.005) is less than the significance level (0.01), we have strong evidence to reject the null hypothesis.

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