Question

Suppose that we are interested in determining whether cats tend to sleep more than 12 hours a day on average let denote the true number of hours that cats sleep each night. You observe 32 cats for a day and record the number of hours that each of the cats slept. The average of your samples 13.1 hours and the sample standard deviation of your sample is 2.8 hours. Your confidence intervalis (a) Construct a 99% confidence interval for hours per day on average ). This means you are 99% confidence that cats sleep between (b) You are now interested in testing the . This means that you pothesis described above at the 0.05 significance level. The test statistic of your hypothesis tests 3 the null hypothesis at the 0.05 significance level, and you conclude that on average cats sleep- and the p-value of your hypothesis 3 2 hours a day.

Answer 1

a) df = 32 - 1 = 31

At 99% confidence level, the critical value is t^* = 2.744

The 99% confidence interval is

$\large \bar x \pm t^{* } * \frac{s}{\sqrt n}$

$\large = 13.1 \pm 2.744 * \frac{2.8}{\sqrt {32}}$

$\large = 13.1 \pm 1.358$

$\large = 11.742, 14.458$

The confidence interval is (11.742, 14.458).

We are 99% confidence that cats sleep between 11.472 and 14.458 hours per day on average.

b) The test statistic is

$\large t = \frac{\bar x - \mu}{s/\sqrt n}$

  $\large = \frac{13.1 - 12}{2.8/\sqrt {32}}$

$\large = 2.22$

P-value = P(T > 2.22)

= 1 - P(T < 2.22)

= 1 - 0.9831

= 0.0169

Since the P-value is less than the significance level (0.0169 < 0.05), so we should reject the null hypothesis.

We reject the null hypothesis at the 0.05 significance level, and we conclude that on average cats sleep more than 12 hours a day.