Question

Suppose the food label on a cookie bag states that there is at most 2 grams of saturated fat in a single cookie. In a sample of 35 cookies, it is found that the mean amount of saturated fat per cookie is 2.1 grams. Assume that the sample standard deviation is 0.3 gram. At 0.05 significance level, can we reject the claim on food label

Answer 1

Answer:

$t=\frac{2.1-2}{\frac{0.3}{\sqrt{35}}}=1.972$  

$p_v =P(t_{34}>1.972)=0.0284$  

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 reject the null hypothesis, so we can conclude that the true mean is significantly higher than 2 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=2.1$ represent the sample mean

$s=0.3$ represent the sample standard deviation

$n=35$ sample size  

$\mu_o =2$ 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 is at least 2 grams, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 2$  

Alternative hypothesis:$\mu > 2$  

Since we don't know the population deviation, 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:  

$t=\frac{2.1-2}{\frac{0.3}{\sqrt{35}}}=1.972$  

P-value  

We calculate the degrees of freedom given by:

$df = n-1 = 35-1 = 34$

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

$p_v =P(t_{34}>1.972)=0.0284$  

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 reject the null hypothesis, so we can conclude that the true mean is significantly higher than 2 at 5% of signficance.