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Answer:
The 95% confidence interval for the difference of means is (7.67, 16.33).
The estimate is Md = 12.
The standard error is sM_d = 2.176.
Step-by-step explanation:
We have to calculate a 95% confidence interval for the difference between means.
The sample 1 (this year's sales), of size n1=36 has a mean of 53 and a standard deviation of 12.
The sample 2 (last year's sales), of size n2=49 has a mean of 41 and a standard deviation of 6.
The difference between sample means is Md=12.
The estimated standard error of the difference between means is computed using the formula:
The degrees of freedom are:
The critical t-value for a 95% confidence interval and 83 degrees of fredom is t=1.989.
The margin of error (MOE) can be calculated as:
Then, the lower and upper bounds of the confidence interval are:
The 95% confidence interval for the difference of means is (7.67, 16.33).
Answer:
4.75% probability that the line pressure will exceed 1000 kPa during any measurement
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the probability that the line pressure will exceed 1000 kPa during any measurement
This is 1 subtracted by the pvalue of Z when X = 1000. So
has a pvalue of 0.9525
1 - 0.9525 = 0.0475
4.75% probability that the line pressure will exceed 1000 kPa during any measurement