Q 14RP. - Elementary Statistics

Question

Blood Glucose Level. In the article "Drinking Glucose Improves Listening Span in Students Who Miss Breakfast" (Educational Research, Vol. 43, No. 2, pp. $201-207$). authors N. Morris and P. Sarll explored the relationship between students who skip breakfast and their performance on a number of cognitive tasks. According to their findings, blood glucose levels in the morning, after a $9-$hour fast, have a mean of $4.60\mathrm{mmol}/\mathrm{L}$ with a standard deviation of $0.16\mathrm{mmol}/\mathrm{L}$. (Note; $mmol/L$ is an abbreviation of millimoles/liter. which is the world standard unit for measuring glucose in the blood.)

a. Determine the sampling distribution of the sample mean for samples of size $60$

b. Repeat part (a) for samples of size $120$

c. Must you assume that the blood glucose levels are normally distributed to answer parts (a) and (b)? Explain your answer.

Step-by-Step Solution

Answer

Part (a) the sampling mean distribution is normal, with a mean of $4.60\mathrm{mmol}/\mathrm{l}$and S.D. $0.021\mathrm{mmol}/\mathrm{l}$

Part (b) the sampling mean distribution is normal, with a mean of $4.60\mathrm{mmol}/\mathrm{l}$ and S.D. $0.015\mathrm{mmol}/\mathrm{l}$

Part (c) No.

Part (a) Step 1: Given information

In the morning, the population mean blood glucose level is $\mu =4.60\mathrm{mmol}/\mathrm{l}$, and the population S.D of blood glucose levels is $\sigma =0.16\mathrm{mmol}/\mathrm{l}$

Part (a) Step 2: Concept

The formula used: Standard deviation ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Part (a) Step 3: Calculation

Here is the sample size $n=60$

As a result, because the sample size of $60$ is more than $30$, we can consider the sample to be substantial.

By the application of CLT,

The sample mean $\overline{x}$ follows roughly. Given a normal distribution.

Mean ${\mu }_x=\mu =4.60$ and Standard deviation ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

$$\begin{gathered} =\frac{0.16}{\sqrt{60}} \\ =0.021 \end{gathered}$$

As a result, the sampling means distribution is normal, with a mean of $4.60\mathrm{mmol}/\mathrm{l}$ and S.D. $0.021\mathrm{mmol}/\mathrm{l}$

Part (b) Step 1: Calculation

The sample size is shown here. $n=120$

As a result, we can consider the sample to be a large sample because the sample size of $120$ is significantly above $30$

 By the application of CLT,

Sample average With $\overline{x}$, the distribution is roughly Normal.

Mean and standard deviation are ${\mu }_x=\mu =4.60$ and ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$ respectively.

$$\begin{gathered} =\frac{0.16}{\sqrt{120}} \\ =0.015 \end{gathered}$$

As a result, the sampling mean distribution is normal, with a mean of $4.60\mathrm{mmol}/\mathrm{l}$ and S.D. $0.015\mathrm{mmol}/\mathrm{l}$

Part (c) Step 1: Calculation

No, for samples of size $60$ and $120$, it is not necessary to assume that blood glucose levels are distributed regularly. Because, regardless of the population distribution, sample means are normally distributed with large samples (i.e., sample size $>30$), sample means are normally distributed with large samples mean ${\mu }_{\overline{x}}=\mu$and S.D. ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$where $\mu = Population mean, \sigma = Population S.D and n= Sample size$