Question

The maker of a certain brand of low-fat cereal bars claims that the average saturated fat content is 0.5 gram. In a random sample of 8 cereal bars of this brand, the saturated fat content was 0.6, 0.7, 0.7, 0.3, 0.4, 0.5, 0.4, and 0.2. Assume the saturated fat content in these bars is normally distributed

a. Find the mean of the sample to three decimals
b. Find the probability that a sample has less saturated fat than the sample mean (your answer to part a). Round to two decimal places.
c. Do you agree with the claim?

Answer 1

Answer:

a) $\bar X= \frac{0.6+0.7+0.7+0.3+0.4+0.5+0.4+0.2}{8}= 0.475$

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

And after replace we got $s= 0.1832$

We assume that the true mean for this case is $\mu = 0.5$

And we can approximate the distribution of the sample mean with a t distribution with $df = n-1 = 8-1 =7$

$\bar X \sim t(\mu , \frac{s}{\sqrt{n}})$

$P(\bar X <0.5) = P(t_{7} < \frac{0.475-0.5}{\frac{0.1832}{\sqrt{8}}}) = P(t_{7} < -0.386) = 0.36$

c) For this case using a 5% of significance if we find on the t distribution with 7 degrees of freedom two values that accumulates 95% of the middle area we gtot $t_{crit}= \pm 2.365$

And the statistic for this case was -0.386

So then the statistic calculated is on the NON rejection zone of the null hypothesis since is between -2.365 and 2.365 so we don't have problems with the claim

Step-by-step explanation:

Data given: 0.6, 0.7, 0.7, 0.3, 0.4, 0.5, 0.4, and 0.2.

Part a

We can calculate the sample mean with the following formula:

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

And if we replace we got:

$\bar X= \frac{0.6+0.7+0.7+0.3+0.4+0.5+0.4+0.2}{8}= 0.475$

Part b

For this case we can begin calculating the sample mean given by this formula:

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

And after replace we got $s= 0.1832$

We assume that the true mean for this case is $\mu = 0.5$

And we can approximate the distribution of the sample mean with a t distribution with $df = n-1 = 8-1 =7$

$\bar X \sim t(\mu , \frac{s}{\sqrt{n}})$

$P(\bar X <0.475) = P(t_{7} < \frac{0.475-0.5}{\frac{0.1832}{\sqrt{8}}}) = P(t_{7} < -0.386) = 0.36$

Part c

For this case using a 5% of significance if we find on the t distribution with 7 degrees of freedom two values that accumulates 95% of the middle area we gtot $t_{crit}= \pm 2.365$

And the statistic for this case was -0.386

So then the statistic calculated is on the NON rejection zone of the null hypothesis since is between -2.365 and 2.365 so we don't have problems with the claim