5) a random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. assuming…

Explanation:

Step1: State hypotheses

$H_0: \mu = 39.9$ (null hypothesis) $H_1: \mu \neq 39.9$ (alternative hypothesis)

Step2: Calculate test statistic

Formula: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ Substitute values: $\bar{x}=40.5$, $\mu=39.9$, $\sigma=1.6$, $n=100$ $z = \frac{40.5 - 39.9}{\frac{1.6}{\sqrt{100}}} = \frac{0.6}{\frac{1.6}{10}} = \frac{0.6}{0.16} = 3.75$

Step3: Find critical value

For $\alpha=0.05$ (two-tailed), critical $z$-values are $\pm 1.96$

Step4: Compare and make decision

$3.75 > 1.96$, so reject $H_0$

Answer:

There is sufficient evidence at the 0.05 significance level to reject the claim that the mean circumference of all pumpkins is equal to 39.9 cm.