What is 66 2/3% as a decimal? NOW
2 answers:
You might be interested in
Answer:
a) X~N(μ;σ²)
b) [87.695 ; 89.464]
c) and d) check the explanation below
Step-by-step explanation:
Hello!
The objective of this experiment is to test if the claim of the company is correct and the 3-ounce bags of candies contain over 90 pieces of candy.
For this, a sample of 26 bags was taken and the number of candies per bag was counted.
The study variable is X: number of candy pieces in a 3 ounces bag.
a)
Using the sample data I've made a Shapiro-Wilks test,
H₀: X has a normal distribution
H₁: X doesn't have a normal distribution
α: 0.05
p-value: 0.5675
Since the p-value is greater than the significance level, we can assume that the study variable has normal distribution:
X~N(μ;σ²)
b)
To make this confidence interval, considering I only have sample data, I choose to use the Student t statistic.
Sample mean X[bar]= 88.58
Sample standard deviation S= 2.19
The formula for the interval is:
X[bar] ± * (S/√n)
88.58 ± 2.060 * (2.19/√26)
[87.695 ; 89.464]
c)
With a confidence level of 95%, you'd expect that the interval [87.69; 89.46] contains the value of the population mean of the number of candy pieces in a 3 ounces bag.
d)
The SunDrop Candies' claim is that their 3 ounces bag contains over 90 pieces of candy. Symbolically: μ > 90
Since their claim would lead to a one-tailed hypothesis test and the calculated confidence interval is two-tailed, you cannot use it to decide whether or not the company's claim is true. But, seeing as the two bonds of the interval are below 90 you could think that the company's claim is not correct. Of course this is just an idea, you must perform a statistical test to test if the company is wrong or not and, of course, have a statistical backing of your conclusions.
I hope it helps!