Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Data. Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time.

The 99% confidence interval for the mean germination time is (12.3, 19.3).

Step-by-step explanation:

The question is incomplete:

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.

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).

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted