Question

Hello everyone, I tried calculating using t and z score and didn't get the answer.

Question 30 0 out of 6 points Which of the following accurately represents a 95% confidence interval for a sample of 5 individuals whose blood glucose levels were 98, 111, 105, 120, and 147? (78.9, 153.5) (153.5, 78.9) (78.9, 153.5) Selected Answer: Answers: (63.4, 169.0) (169.0, 63.4)

Answer 1

Solution: We are required to find the 95% confidence interval for the population mean.

The 95% confidence interval for the population mean is:

Where:

$\bar{x}=116.2$ is the sample mean of given 5 observations

$t_{\frac{0.05}{2}}=2.776$ is the critical value at 0.05 significance level for $df=n-1=5-1=4$

$s=19.018$ is the sample standard deviation of given 5 observations

$n=5$ is the number of individuals in the given sample

$\therefore 116.2\pm 2.776\left ( \frac{19.018}{\sqrt{5}} \right )$

$116.2\pm 23.6$

$\left [ 116.2- 23.6,116.2+ 23.6 \right ]$

$\left [ 92.6,139.8 \right ]$

Therefore, the 95% confidence interval for the population mean is $\left ( 92.6,139.8 \right )$

But none of the given options is what we got.

There is a mistake with the given correct answer $\boldsymbol{\left ( 63.4,169.0 \right )}$ .

Let me explain it:

The answer given above has been calculated using the below formula:

$\bar{x}\pm t_{\frac{0.05}{2}} \times s$

$116.2\pm 2.776 \times 19.018$

$116.2\pm 52.8$

$\left ( 116.2- 52.8,116.2+ 52.8 \right )$

$\left ( 63.4,169.0 \right )$

So using the formula $\bar{x}\pm t_{\frac{0.05}{2}} \times s$ , we got exactly what has been shown the correct answer. But the formula used is wrong because they have forgotten to divide standard deviation by $\sqrt{n}$ . That is the reason you were unable to figure out the correct answer.

Hope this helps!