Question

A maker of a certain brand of low-fat cereal barsclaims that the average saturated fat content is 0.5gram. In a random sample of 8 cereal bars of thisbrand, the saturated fat content was 0.6, 0.7, 0.7, 0.3,0.4, 0.5, 0.4, and 0.2. Would you agree with the claim?Assume a normal distribution.

Answer 1

Yes, it is actually

Explanation:

Answer 2

`t=(0.475-0.5)/((0.183)/(√(8)))=-0.386`

`p_v =2*P(t_((7))<-0.386)=0.711`

If we compare the p value and the significance level assumed
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance. So then the claim makes sense

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

In order to calculate the mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}` (3)

The mean calculated for this case is
`\bar X=0.475`

The sample deviation calculated
`s=0.183`

We need to conduct a hypothesis in order to check if the true mean is 0.5 or no, the system of hypothesis would be:

Null hypothesis:
`\mu = 0.5`

Alternative hypothesis:
`\mu \\eq 0.5`

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(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=(0.475-0.5)/((0.183)/(√(8)))=-0.386`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=8-1=7`

Since is a two sided test the p value would be:

`p_v =2*P(t_((7))<-0.386)=0.711`

Conclusion

If we compare the p value and the significance level assumed
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance. So then the claim makes sense