Answer to Question #144013 in Statistics and Probability for J

Question

You want to impress your family and you are wondering if you could find a farm where you could be sure to get very big pumpkins, like 25 pounds on average. A friend shows you the 4 pumpkins she got at the SuperPumpkins Farm, and you are able to calculate that the mean weight of this sample of 4 pumpkins to be 27 pounds. To find out the true population mean weight at the SuperPumpkins Farm, you decide to build a 99% confidence interval with this sample data from your friend, assuming that the population standard deviation should be 6 pounds (i.e. typical for pumpkins, the same as the Huckleberry Farm). Based on this confidence interval you built, do you feel confident that the SuperPumpkins Farm will have very big pumpkin of around 25 pounds on average. round to 2 decimals, !

Answer 1

We need to construct the 99% confidence interval for the population mean $\mu.$

The following information is provided: $\bar{x}=27, \sigma=27, n=4.$

The critical value for $\alpha=0.01$ is $z_c=z_{1-\alpha/2}=2.576.$ The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\cdot\sigma/\sqrt{n}, \bar{x}+z_c\cdot\sigma/\sqrt{n})=$$

$$=(27-2.576\cdot6/\sqrt{4}, 27+2.576\cdot6/\sqrt{4})=$$

$$=(19.273, 34.727)\approx(19.27, 34.73)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $19.27<\mu< 34.73,$ which indicates that we are 99% confident that the true population mean $\mu$  is contained by the interval (19.27, 34.73).

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq25$

$H_1:\mu<25$

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.01,$ and the critical value for a left-tailed test is $z_c=-2.33.$  

The rejection region for this left-tailed test is $R=\{z:z<-2.33\}$  

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{27-25}{6/\sqrt{4}}=\dfrac{2}{3}\approx0.667$$

Since it is observed that $z=0.667>-2.33=z_c,$  z = 0.667 it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 25, at the 0.01 significance level.