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.

Short 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.

Step-by-step solution

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 $\mu =Populationmean,\sigma =PopulationS.Dandn=Samplesize$