MATH SOLVE

4 months ago

Q:
# 0 A 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. Would you agree with the claim? Assume a normal distribution

Accepted Solution

A:

Answer:The first step is to determine the average
[tex]x= \frac{0,6+0,7+ 0,7+0,3+0,4+ 0,5+ 0,4 +0,2}{8}\\\ x= 0,475[/tex]The exercise says it’s a normal distribution: (n=8)
[tex]s^{2}= \frac{1}{n-1} ((0,6.x)^2+2(0,7-x)^2+(0.3-x)^2+2(0.4-x)^2+(0.5-x)^2+(0.2-x)^2)\\\ \frac{1}{7}*0,235 = 0,0336 \\\\ s= 0183[/tex]According to the exercise, the mean is equal to 0,5 then the value of t of the distribution can be obtained
[tex]t= \frac{x-u}{\frac{s}{\sqrt{n} }}\\\ t= \frac{0,475 - 0,5}{0,183}\\\ t= -0,3860[/tex]The variable t has 7 grade to liberty, we calculate the p-value as:
[tex]2* P(t < - 0,3860)= \\\ 2* 0,3554 = 0,7108[/tex]This value is very high, therefore the hypothesis is not rejected