Answer to Question #130614 in Statistics and Probability for Abdul Rahman Md Yunus

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

Answer 1

ANSWER 1)

Let us consider the null hypothesis H₀ : $\mu = 300$ against the claim of the researcher H₁ : $\mu < 300$ .

Test Criteria : Reject the null hypothesis H₀ if Z_calc < Z$\alpha$

Given that,

Sample quantity N = 50

Sample mean U = 296

Standard deviation $\sigma = 5$

Considering the population mean $\mu = 300$

We can find Z_calc = $\frac{U-\mu}{\frac{\sigma}{\sqrt N}}$

Z_calc = $\frac{296-300}{\frac{5}{\sqrt 50}} = -5.6568$

Since it is a left tailed test we will have to find the value of Z_(1-$\alpha$₎

Z_(1-$\alpha$_{) =} Z_{(1-0.05) =} -1.6449

Hence, Z_calc < Z_(1-$\alpha$₎ as -5.6568 < -1.6449

So we can reject the null hypothesis and say that the claim of the researcher is true.

ANSWER 2)

The 95% confidence interval for a left-tailed test (when level of significance = $\alpha$ ) is :

$(-\infin, U - Z_{1-\alpha}*\frac{\sigma}{\sqrt n}]$

$(-\infin, 296 - (-1.6449)*\frac{5}{\sqrt 50}]$

$(-\infin, 297.1631]$

95% confidence interval for μ is $(-\infin, 297.1631]$

But in this case since the population mean μ cannot be $-\infin$ since it is in terms of weight so we can reduce the interval to [0, 297.1631]