Mathematics, 24.12.2019 23:31 woodfordmaliky

Afederal agency responsible for enforcing laws governing weights and measures routinely inspects packages to determine whether the weight of the contents is at least as great as that advertised on the package. a random sample of 18 containers whose packaging states that the contents weigh 8 ounces was drawn. the contents were weighed and the results follow. can we conclude at the 1% significance level that on average the containers are mislabeled?

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Answer from: shongmadi77

$t=\frac{7.96-8}{\frac{0.085}{\sqrt{18}}}=-1.997$

$p_v =P(t_{(17)}$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the population mean is not significantly lower than 8 ounces at 1% of significance.

Step-by-step explanation:

Assuming thsi info: "The sample mean was 7.96 ounces and the sample standard deviation is 0.085 ounces."

Previous concepts and data given

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=7.96$ represent the sample mean

s=0.085 represent the sample standard deviation

n=18 represent the sample selected

$\alpha=0.01$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 8 ounces, the system of hypothesis would be:

Null hypothesis: $\mu \geq 8$

Alternative hypothesis: $\mu < 8$

If we analyze the size for the sample is ><30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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:

$t=\frac{7.96-8}{\frac{0.085}{\sqrt{18}}}=-1.997$

P-value

The first step is calculate the degrees of freedom, on this case:

df=n-1=18-1=17