Using the z-distribution, it is found that since the p-value is less than 0.05, there is evidence that the mean amount of cereal in each box is different from 16 ounces at 0.05 significance.
At the null hypothesis, it is tested if the mean is not different to 16 ounces, that is:
[tex]H_0: \mu = 16[/tex]
At the alternative hypothesis, it is tested if the mean is different, hence:
[tex]H_1: \mu \neq 16[/tex]
The test statistic is:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which:
The parameters for this problem are:
[tex]\overline{x} = 15.75, \mu = 16, \sigma = 1.46, n = 150[/tex].
Hence:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{15.75 - 16}{\frac{1.46}{\sqrt{150}}}[/tex]
z = -2.1
Using a z-distribution calculator, for a two-tailed test, as we are testing if the mean is different of a value, with z = -2.1, the p-value is of 0.0357.
Since the p-value is less than 0.05, there is evidence that the mean amount of cereal in each box is different from 16 ounces at 0.05 significance.
More can be learned about the z-distribution at https://brainly.com/question/16313918
#SPJ1