228k views
3 votes
A canned soup company would like to test if the average fill in their soup cans is different than their claimed amount of 16 ounces. They take a simple random sample of 101 cans of soup. The sample yields a sample mean and sample standard deviations of 15.8 ounces and LaTeX: s = 0.9 \: ouncess = 0.9 o u n c e s.

User Nulltoken
by
8.2k points

1 Answer

3 votes

Answer:

The average fill in the soup cans is different than the claimed amount of 16 ounces.

Explanation:

A one-sample t-test can be used to test whether the population mean is significantly different from some hypothetical value of 16 ounces or not.

The hypothesis is:

H₀: The average fill in the soup cans is 16 ounces, i.e. µ = 16.

Hₐ: The average fill in the soup cans is different from 16 ounces, i.e. µ ≠ 16.

Assume that the significance level of the test is α = 0.05.

The t-statistic is given by,


t=(\bar x-\mu)/(s/√(n))

Here


\bar x = sample mean = 15.8 ounces

s = sample standard deviation = 0.9 ounces

n = sample size = 101.

Compute the value of the test statistic as follows:


t=(\bar x-\mu)/(s/√(n))=(15.8-16)/(0.9/√(101))=-2.23

The calculated t-statistic is, t = -2.23 which follows t-distribution with (n-1) = 100 degrees of freedom.

Decision rule:

If the p-value of the test statistic is less than the significance level α then the null hypothesis will be rejected. And if the p-value is more than α then the null hypothesis will not be rejected.

The p-value is,

p-value = 0.028.

Use the p-value from t-score calculator.

The calculated p-value, p = 0.028 < α = 0.05.

The null hypothesis will be rejected at 5% level of significance.

Conclusion:

The average fill in the soup cans is different than the claimed amount of 16 ounces.

User Lukewm
by
8.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.