Frank has 24 pennies, 62 nickels, 55 dimes, 16 quarters, and 19 fifty-cent pieces. How much money does he have?
1 answer:
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Answer:
Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131
Rejection Zone: or
In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.
Step-by-step explanation:
Previous concepts and data given
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
represent the sample mean
represent the sample standard deviation
n=16 represent the sample selected
significance level
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
Critical values
On this case since we have a bilateral test we need to critical values. We need to use the t distribution with degrees of freedom. The value for
and
so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:
"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"
And we got
So the decision on this case would be:
Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131
Rejection Zone: or
Conclusion
In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.