Express 20p of a fraction of 5, Write 0.065 as a fraction
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Answer:
We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.
Step-by-step explanation:
We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.
A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.
Let = <u><em>mean weight of coffee in its containers.</em></u>
SO, Null Hypothesis, :
32 ounces {means that the mean weight of coffee in its containers is at least 32 ounces}
Alternate Hypothesis, :
< 32 ounces {means that the mean weight of coffee in its containers is less than 32 ounces}
The test statistics that would be used here <u>One-sample t-test statistics</u> as we don't know about population standard deviation;
T.S. = ~
where, = sample mean weight = 31.8 ounces
s = sample standard deviation = 0.48 ounces
n = sample of containers = 15
So, <u><em>the test statistics</em></u> = ~
= -1.614
The value of t test statistics is -1.614.
<u>Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.</u>
Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.
Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.