Question

10. A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5cm. Assuming that the population standard deviation is known to be 1.6cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9cm

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=39.9$

$H_0:\mu\not=39.9$

This corresponds to a two-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 $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z: |z| > 1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{40.5-39.9}{1.6/\sqrt{100}}=3.75$$

Since it is observed that $|z| = 3.75 > 1.96=z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(Z>3.75)=0.000177,$ and since $p=0.000177<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is evidence to claim that the population mean $\mu$

is different than $39.9,$ at the $\alpha = 0.05$ significance level.