Question

Suppose that the quality control manager for a cereal manufacturer wants to ensure that bags of cereal are being filled correcty The equipment is calibrated to fill bags with a mean of 16 oz of cereal with a standard deviation of 0.2 oz. The quality control inspector selects a random sample of 45 boxes and finds that the mean amount of cereal for these boxes is 16.03 oz. He uses this data to conduct a Hi: ? 16, where ? is the mean amount of cereal in each box. He calculates a z-score of 1.01 and a p-value of 0.3125.

Are these results statistically significant at a significance level of 0.05?

O No, these results are not statistically significant because p>0.05,
O No, these results are not statistically significant because p < 0.05.
O Yes, these results are statistically significant because p < 0.05.
O Yes, these results are statistically significant because p > 0.05.

Answer 1

No, these results are not statistically significant because p>0.05,

Explanation:

Data given and notation

`\bar X=16.03` represent the sample mean

`\sigma=0.2` represent the population standard deviation for the sample

`n=45` sample size

`\mu_o =16` represent the value that we want to test

`\alpha=0.05` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the content differs from 16, the system of hypothesis would be:

Null hypothesis:
`\mu =16`

Alternative hypothesis:
`\mu \\eq 16`

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`z=(16.03-16)/((0.2)/(√(45)))=1.006`

P-value

Since is a two-sided test the p value would be:

`p_v =2*P(Z>1.006)=0.3125`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

The best conclusion for this case would be:

No, these results are not statistically significant because p>0.05,