Mathematics, 14.12.2019 00:31 rubincain698

Astore sells "16-ounce" boxes of captain crisp cereal. a random sample of 9 boxes was taken and weighed. the results were the following (in ounces): 15.5 16.2 16.1 15.8 15.6 16.0 15.8 15.9 16.2 assume the weight of cereal in a box is normally distributed. recall: x = 15.9, s = 0.25.
a) the company that makes captain crisp cereal claims that the average weight of its box is at least 16 ounces. use a 0.05 level of significance to test the company's claim. b) some customers believe that the variance of the weight of boxes with captain crisp cereal is more than 0.05 oz^2. use a = 0.10 to test the customers’ claim.

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Answer from: tdahna0403

a) $p_v =P(\chi^2_{8}$

On this case since the p value it's higher than the significance level we FAIL to reject the null hypothesis, and we don't have anough evidence to conclude that the true mean is less than 16.

b) $p_v = P(\chi^2_{8}10)=0.265$

If we compare the p value and the significance level given we see that $p_v \alpha$ so then we have enough evidence to FAIL to reject the null hypothesis that the true population deviation it's not significantly higher than 0.224 oz^2.

Step-by-step explanation:

Part a

Data given and notation

We can calculate the sample mean and standard deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

$\bar X=15.9$ represent the sample mean

s=0.25 represent the standard deviation for the sample

n=9 sample size

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

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

t 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 to be tested

We need to conduct a hypothesis in order to determine if the average weight of its box is at least 16 ounces, the system of hypothesis would be:

Null hypothesis: $\mu \geq 16$

Alternative hypothesis: $\mu < 16$

Compute the test statistic

We don't know the population deviation, so for this case 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".

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

$t=\frac{15.9-16}{\frac{0.25}{\sqrt{9}}}=-1.2$

Now we need to find the degrees of freedom for the t distribution given by:

df=n-1=9-1=8

What do you conclude? Use the p-value approach

Since is a left tailed test the p value would be:

$p_v =P(\chi^2_{8}$

On this case since the p value it's higher than the significance level we FAIL to reject the null hypothesis, and we don't have anough evidence to conclude that the true mean is less than 16.

Part b

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 0.224, so the system of hypothesis are:

H0: $\sigma \leq 0.224$

H1: $\sigma 0.224$

In order to check the hypothesis we need to calculate the statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(9-1) [\frac{0.25}{0.224}]^2 =10$

What is the critical value for the test statistic at an α = 0.01 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 8 degrees of freedom that accumulates 0.1 of the area on the right tail and 0.90 on the left tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.9,8)". And our critical value would be $\chi^2 =13.362$