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sergejj [24]
3 years ago
10

A canned soup company would like to test if the average fill in their soup cans is different than their claimed amount of 16 oun

ces. 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.
Mathematics
1 answer:
xxMikexx [17]3 years ago
5 0

Answer:

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

Step-by-step explanation:

A one-sample <em>t</em>-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:

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

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

Assume that the significance level of the test is <em>α</em> = 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 <em>p</em>-value of the test statistic is less than the significance level  <em>α</em> then the null hypothesis will be rejected. And if the <em>p</em>-value is more than <em>α</em> then the null hypothesis will not be rejected.

The p-value is,

<em>p</em>-value = 0.028.

Use the p-value from <em>t</em>-score calculator.

The calculated p-value, <em>p</em> = 0.028 < <em>α</em> = 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.

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