Answer to Question #130481 in Statistics and Probability for Aznoor

Question

A researcher wishes to test the claim of a particular cereal manufacturer that the mean weight of cereal in the boxes is less than 300g. A sample of 50 boxes yields a sample mean weight of 296g. Assume that the population standard deviation is 5g.



Can we conclude that the claim is true? Test at α = 0.05
Obtain a 95% confidence interval for μ.

Answer 1

The provided sample mean is $\bar X = 296$ and the known population standard deviation is σ=5, and the sample size is n = 50

1) Null and Alternative Hypotheses

$H_{o}:\mu =300$

$H_{a}:\mu <300$

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

2 .Test Statistics

The z-statistic is computed as follows:

$z=\frac{\overline{X}-mean}{\frac{\sigma }{\sqrt{n}}}$

$z=\frac{296-300}{\frac{5 }{\sqrt{50}}}$ = −5.657

Using the P-value approach: P value is the value to the left of -5.657 . That is, the p-value is p = 0 and since p=0<0.05, it is concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 300, at the 0.05 significance level.

confidence interval :

formula : $[\overline{X}-\frac{Z^{*}*\sigma }{\sqrt{n}},\overline{X}+\frac{Z^{*}*\sigma }{\sqrt{n}}]$

We need to construct the 95% confidence interval for the population mean μ. Using Z table or ti84, the critical value for α=0.05 is $z_c = z_{1-\alpha/2} = 1.96$ . The corresponding confidence interval is computed as shown below:

$[296-\frac{1.96*5 }{\sqrt{50}},296+\frac{1.96*5 }{\sqrt{50}}]$

[ 294.614,297.386 ]

Therefore, based on the data provided, the 95% confidence interval for the population mean is 294.614<μ<297.386, which indicates that we are 95% confident that the true population mean μ is contained by the interval (294.614, 297.386)