Question

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.

Answer 1

Answer:

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

Step-by-step 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,  

$t=\frac{\bar x-\mu}{s/\sqrt{n}}$

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:

        $t=\frac{\bar x-\mu}{s/\sqrt{n}}=\frac{15.8-16}{0.9/\sqrt{101}}=-2.23$          

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.