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Answer:
a) The sample mean is of 49 and the sample standard deviation is of 11.7.
b) The range of the true mean at 90% confidence level is of 9.62 hours.
c) The prediction interval, at a 90% confidence level, of it's failure time is between 39.38 hours and 58.62 hours.
Step-by-step explanation:
Question a:
Sample mean:
Sample standard deviation:
The sample mean is of 49 and the sample standard deviation is of 11.7.
b)Determine the range of the true mean at 90% confidence level.
We have to find the margin of error of the confidence interval. Since we have the standard deviation for the sample, the t-distribution is used.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 6 - 1 = 5
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.0.150
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample. So
The range of the true mean at 90% confidence level is of 9.62 hours.
(c)If a seventh sample is tested, what is the prediction interval (90% confidence level) of its failure time.
This is the confidence interval, so:
The lower end of the interval is the sample mean subtracted by M. So it is 49 - 9.62 = 39.38 hours.
The upper end of the interval is the sample mean added to M. So it is 49 + 9.62 = 58.62 hours.
The prediction interval, at a 90% confidence level, of it's failure time is between 39.38 hours and 58.62 hours.
Solution :
Let and
represents the proportions of the seeds which germinate among the seeds planted in the soil containing
and
mushroom compost by weight respectively.
To test the null hypothesis against the alternate hypothesis
.
Let denotes the respective sample proportions and the
represents the sample size respectively.
The test statistic can be written as :
which under follows the standard normal distribution.
We reject at
level of significance, if the P-value
or if
Now, the value of the test statistics = -1.368928
The critical value =
P-value =
= 0.171335
Since the p-value > 0.05 and , so we fail to reject
at
level of significance.
Hence we conclude that the two population proportion are not significantly different.
Conclusion :
There is not sufficient evidence to conclude that the of the seeds that
with the percent of the
in the soil.