Question

You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures: 44 32.8 59.2 31.4 12.7 68.5 84.7 72.5 55.7 Find the 98% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place). 98% C.I.

Answer 1

Answer:

$51.278 -2.896 \frac{22.979}{\sqrt{9}}= 29.096$

$51.278 +2.896 \frac{22.979}{\sqrt{9}}= 73.460$

And the interval would be:

$(29.10 \leq \mu \leq 73.46)$

Step-by-step explanation:

For this problem we have the following dataset given:

44 32.8 59.2 31.4 12.7 68.5 84.7 72.5 55.7

We can find the mean and sample deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$\bar X= 51.278$

$s= 22.979$

The confidence interval for the mean is given by:

$\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

The degrees of freedom are:

$df=n-1= 9-1=8$

The confidence would be 0.98 and the significance $\alpha=0.02$ then the critical value would be:

$t_{\alpha/2}= 2.896$

Ad replacing we got:

$51.278 -2.896 \frac{22.979}{\sqrt{9}}= 29.096$

$51.278 +2.896 \frac{22.979}{\sqrt{9}}= 73.460$

And the interval would be:

$(29.10 \leq \mu \leq 73.46)$