Can somebody solve this for me $1.12 for 9.2 ounces
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Answer:
The 99% confidence interval for the mean germination time is (12.3, 19.3).
Step-by-step explanation:
<em>The question is incomplete:</em>
<em>Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time.</em>
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We start calculating the sample mean M and standard deviation s:
We have to calculate a 99% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=15.8.
The sample size is N=10.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
The degrees of freedom for this sample size are:
df=n-1=10-1=9
The t-value for a 99% confidence interval and 9 degrees of freedom is t=3.25.
The margin of error (MOE) can be calculated as:
Then, the lower and upper bounds of the confidence interval are:
The 99% confidence interval for the mean germination time is (12.3, 19.3).